Differential Equations Notes

Higher Order Differential Equations

General Form:

\[ a_n(x) \frac{d^n y}{dx^n} + a_{n-1}(x) \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1(x) \frac{dy}{dx} + a_0(x)y = g(x) \]

To solve a differential equation of degree \(n\), \(n\) initial conditions must be known:

\[ \begin{aligned} y(x_0) &= y_0 \\ y'(x_0) &= y_1 \\ y''(x_0) &= y_2 \\ &\vdots \\ y^{(n-1)}(x_0) &= y_{n-1} \end{aligned} \]

Types of Solutions

  • Initial Value Problem (IVP) – with respect to time
  • Boundary Value Problem (BVP) – with respect to space

Existence of Unique Solution

A unique solution exists if:

  • All functions \(a_n(x), a_{n-1}(x), \ldots, a_0(x), g(x)\) are continuous on an interval \(I\).
  • \(a_n(x) \ne 0\) for all \(x\) in the interval \(I\).

Differential Operator

Define the differential operator:

\(D = \frac{d}{dx}\), \(D^2 = \frac{d^2}{dx^2}\), etc.

Then the general homogeneous differential equation becomes:

\[ a_n(x) D^n(y) + a_{n-1}(x) D^{n-1}(y) + \cdots + a_1(x) D(y) + a_0(x) y \]

This can be written as:

\[ \left[ a_n(x) D^n + a_{n-1}(x) D^{n-1} + \cdots + a_1(x) D + a_0(x) \right] y \]

Define:

\(L = a_n(x) D^n + a_{n-1}(x) D^{n-1} + \cdots + a_0(x)\)

So the equation becomes:

\(Ly = g(x)\)

Here, \(L\) is called the \(n^\text{th}\)-order differential operator or polynomial operator.

Second Order Differential Equations

To solve, at least two initial conditions are needed:

  • Initial Value Problem (IVP): \(y(x_0) = y_0\), \(y'(x_0) = y_1\)

  • Boundary Value Problem (BVP): Examples of conditions (choose any two as required):

\[ \begin{aligned} y(a) &= y_0, \quad y(b) = y_1 \\ y'(a) &= y_0, \quad y'(b) = y_1 \\ y''(a) &= y_0, \quad y''(b) = y_1 \end{aligned} \]

Types of Solutions (for BVPs)

  • Unique solution
  • No solution
  • Infinitely many solutions

Homogeneous Equations

A differential equation is homogeneous if \(Ly = 0\).

If \(y = f(x)\) is a solution, then \(L[f(x)] = 0\).

Also, for any scalar \(\alpha\): \(L[\alpha f(x)] = \alpha L[f(x)] = \alpha \cdot 0 = 0\)

So any scalar multiple of a solution is also a solution.

Superposition Principle

The superposition principle states that if \(y_1, y_2, \ldots, y_k\) are solutions to a homogeneous linear differential equation, then any linear combination of these solutions, \(y = c_1 y_1 + c_2 y_2 + \cdots + c_k y_k\), is also a solution, where \(c_1, c_2, \ldots, c_k\) are arbitrary constants.

\(L(c_1 y_1 + c_2 y_2 + \cdots + c_k y_k) = 0\)

Linear Dependence / Independence

  • Linearly dependent solutions are not valid for forming the general solution.
  • Only linearly independent solutions form a fundamental set.

To check dependence for two functions \(f_1(x)\) and \(f_2(x)\):

\(c_1 f_1(x) + c_2 f_2(x) = 0\) for all \(x\) in the interval, where not both \(c_1, c_2\) are zero. If the ratio \(\frac{f_1(x)}{f_2(x)}\) is constant, the functions are linearly dependent.

Wronskian Method

Given functions \(f_1, f_2, \ldots, f_n\) that are at least \(n-1\) times differentiable, the Wronskian is:

\[ W(f_1, f_2, \ldots, f_n) = \begin{vmatrix} f_1 & f_2 & \cdots & f_n \\ f_1' & f_2' & \cdots & f_n' \\ \vdots & \vdots & \ddots & \vdots \\ f_1^{(n-1)} & f_2^{(n-1)} & \cdots & f_n^{(n-1)} \end{vmatrix} \]

  • If \(W(f_1, f_2, \ldots, f_n) = 0\) for all \(x\) in the interval, the functions are linearly dependent.
  • If \(W(f_1, f_2, \ldots, f_n) \ne 0\) for at least one point in the interval, the functions are linearly independent.

Non-Homogeneous Equations

General form: \(Ly = g(x)\)

Associated homogeneous equation: \(Ly = 0\)

If \(y = f(x)\) is a particular solution, then \(L[f(x)] = g(x)\).

Note that \(L[\alpha f(x)] = \alpha g(x) \ne g(x)\) (unless \(\alpha=1\)) ⇒ Non-homogeneous equations are not scalable by arbitrary constants in the same way homogeneous equations are.

  • Solution to the non-homogeneous equation: Particular solution (\(y_p\))
  • Solution to the homogeneous equation: Complementary solution (\(y_c\))

So, the general solution of a non-homogeneous linear differential equation is the sum of its complementary solution and any particular solution:

\(y = y_c + y_p\)

Where \(y_c\) is the general solution to the associated homogeneous equation \(Ly=0\), and \(y_p\) is any specific solution to \(Ly=g(x)\).

Superposition Principle in Non-Homogeneous Case

If \(g(x)\) can be written as a sum of functions, say \(g(x) = g_1(x) + g_2(x) + \dots + g_k(x)\), and \(y_{p_i}\) is a particular solution to \(Ly = g_i(x)\) for each \(i=1, \dots, k\), then a particular solution to \(Ly = g(x)\) is the sum of these particular solutions:

\(y_p = y_{p_1} + y_{p_2} + \dots + y_{p_k}\).

Homogeneous Linear Equations with Constant Coefficients

General form: \(ay'' + by' + cy = 0\)

Method of Solution: Auxiliary Equation

Assume a solution of the form \(y = e^{mx}\). Then \(y' = me^{mx}\) and \(y'' = m^2 e^{mx}\).

Substituting these into the differential equation: \(am^2 e^{mx} + bm e^{mx} + c e^{mx} = 0\) Factor out \(e^{mx}\): \(e^{mx}(am^2 + bm + c) = 0\)

Since \(e^{mx} \ne 0\), we must have: \(am^2 + bm + c = 0\) This is called the auxiliary equation (or characteristic equation). Solving this quadratic equation for \(m\) yields two roots, \(m_1\) and \(m_2\).

Types of Roots and Corresponding Solutions:

  1. Real and Distinct Roots (\(m_1 \ne m_2\)): Discriminant \(\Delta = b^2 - 4ac > 0\). The general solution is \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\).

  2. Real and Equal Roots (\(m_1 = m_2 = m\)): Discriminant \(\Delta = b^2 - 4ac = 0\). The general solution is \(y = c_1 e^{mx} + c_2 x e^{mx}\).

  3. Complex Conjugate Roots (\(m_1 = \alpha + i\beta\), \(m_2 = \alpha - i\beta\)): Discriminant \(\Delta = b^2 - 4ac < 0\). Using Euler’s Formula (\(e^{i\theta} = \cos\theta + i\sin\theta\)): The general solution is \(y = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))\).

Check/verify solutions after finding them.

This is a homogeneous linear second-order differential equation with constant coefficients.

Step 1: Assume a Solution Form

Assume a solution of the form \(y = e^{mx}\). Then we find its derivatives: \(y' = me^{mx}\) \(y'' = m^2e^{mx}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ 2(m^2e^{mx}) - 5(me^{mx}) - 3(e^{mx}) = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(2m^2 - 5m - 3) = 0 \] The auxiliary (or characteristic) equation is: \[ 2m^2 - 5m - 3 = 0 \]

Step 3: Solve the Auxiliary Equation

This is a quadratic equation. We can solve it by factoring or using the quadratic formula. By factoring: We look for two numbers that multiply to \((2)(-3) = -6\) and add up to \(-5\). These numbers are \(-6\) and \(1\). So, rewrite the middle term: \(2m^2 - 6m + m - 3 = 0\) Factor by grouping: \(2m(m - 3) + 1(m - 3) = 0\) \((2m + 1)(m - 3) = 0\) This gives two distinct real roots: \[ 2m + 1 = 0 \implies 2m = -1 \implies m_1 = -\frac{1}{2} \] \[ m - 3 = 0 \implies m_2 = 3 \]

Step 4: Write the General Solution based on the Roots

For a homogeneous linear second-order differential equation with two distinct real roots \(m_1\) and \(m_2\), the general solution is of the form: \[ y = c_1 e^{m_1 x} + c_2 e^{m_2 x} \] Substitute \(m_1 = -\frac{1}{2}\) and \(m_2 = 3\): \[ y = c_1 e^{-\frac{1}{2}x} + c_2 e^{3x} \] where \(c_1\) and \(c_2\) are arbitrary constants.

Thus, the general solution to the differential equation is \(y = c_1 e^{-\frac{1}{2}x} + c_2 e^{3x}\).

This is a homogeneous linear second-order differential equation with constant coefficients.

Step 1: Assume a Solution Form

Assume a solution of the form \(y = e^{mx}\). Then we find its derivatives: \(y' = me^{mx}\) \(y'' = m^2e^{mx}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ m^2e^{mx} - 10(me^{mx}) + 25(e^{mx}) = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 - 10m + 25) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 - 10m + 25 = 0 \]

Step 3: Solve the Auxiliary Equation

This is a quadratic equation. We can solve it by factoring, recognizing it as a perfect square trinomial: \[ (m-5)^2 = 0 \] This gives a repeated real root: \[ m = 5 \]

Step 4: Write the General Solution based on the Roots

For a homogeneous linear second-order differential equation with a repeated real root \(m\), the general solution is of the form: \[ y = c_1 e^{mx} + c_2 x e^{mx} \] Substitute \(m = 5\): \[ y = c_1 e^{5x} + c_2 x e^{5x} \] where \(c_1\) and \(c_2\) are arbitrary constants.

Thus, the general solution to the differential equation is \(y = c_1 e^{5x} + c_2 x e^{5x}\).

This is a homogeneous linear second-order differential equation with constant coefficients.

Step 1: Assume a Solution Form

Assume a solution of the form \(y = e^{mx}\). Then we find its derivatives: \(y' = me^{mx}\) \(y'' = m^2e^{mx}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ m^2e^{mx} + 4(me^{mx}) + 7(e^{mx}) = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 + 4m + 7) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 + 4m + 7 = 0 \]

Step 3: Solve the Auxiliary Equation

This is a quadratic equation. We will use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=1\), \(b=4\), \(c=7\). \[ m = \frac{-(4) \pm \sqrt{(4)^2 - 4(1)(7)}}{2(1)} \] \[ m = \frac{-4 \pm \sqrt{16 - 28}}{2} \] \[ m = \frac{-4 \pm \sqrt{-12}}{2} \] Simplify the square root of \(-12\): \(\sqrt{-12} = \sqrt{4 \cdot (-3)} = 2\sqrt{3}i\). \[ m = \frac{-4 \pm 2\sqrt{3}i}{2} \] \[ m = \frac{-4}{2} \pm \frac{2\sqrt{3}i}{2} \] \[ m = -2 \pm \sqrt{3}i \] The roots are complex conjugates of the form \(m = \alpha \pm i\beta\), where \(\alpha = -2\) and \(\beta = \sqrt{3}\).

Step 4: Write the General Solution based on the Roots

For a homogeneous linear second-order differential equation with complex conjugate roots \(m = \alpha \pm i\beta\), the general solution is of the form: \[ y = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x)) \] Substitute \(\alpha = -2\) and \(\beta = \sqrt{3}\): \[ y = e^{-2x}(c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x)) \] where \(c_1\) and \(c_2\) are arbitrary constants.

Thus, the general solution to the differential equation is \(y = e^{-2x}(c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x))\).

Behavior of Solutions based on Roots:

  • \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\): This solution generally does not pass through zero multiple times (at most once).
  • \(y = c_1 e^{mx} + c_2 x e^{mx}\): This solution may pass through zero.
  • \(y = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))\): This solution oscillates.
    • If \(\alpha < 0\), it is a damped oscillation.
    • If \(\alpha = 0\), it is an undamped oscillation (occurs when the \(y'\) term is not present in the original equation, i.e., \(b=0\)).

For Equal Roots of Multiplicity R:

If an auxiliary equation has a root \(m\) with multiplicity \(R\), then the corresponding linearly independent solutions are: \(e^{mx}, x e^{mx}, x^2 e^{mx}, \ldots, x^{R-1} e^{mx}\).

Undetermined Coefficients (Method)

This method is used to find a particular solution \(y_p\) for non-homogeneous linear differential equations with constant coefficients, \(Ly = g(x)\), where \(g(x)\) is a sum of terms of the following types: polynomials, exponentials, sines, and cosines.

Annihilator Approach:

A differential operator \(A\) is an annihilator for a function \(g(x)\) if \(A[g(x)] = 0\). Differential operators can be factored, and their order does not matter (they are commutative).

Function \(g(x)\) Annihilator \(A\)
\(c_0, c_1 x, \ldots, c_n x^n\) \(D^{n+1}\)
\(c e^{ax}, c x e^{ax}, \ldots, c x^n e^{ax}\) \((D-a)^{n+1}\)
\(c e^{ax} \cos(\beta x), c e^{ax} \sin(\beta x)\), \((D^2 - 2\alpha D + (\alpha^2 + \beta^2))^{n+1}\)
\(c x^k e^{ax} \cos(\beta x)\), etc. (\(k \le n\)) (For \(n=0\) means functions without \(x\) multiplier)

If \(g(x)\) is a sum of functions, say \(g(x) = g_1(x) + g_2(x)\), and \(A_1\) is the annihilator for \(g_1(x)\) and \(A_2\) is the annihilator for \(g_2(x)\), then \(A_1 A_2\) (or \(A_2 A_1\)) is an annihilator for \(g_1(x) + g_2(x)\).

This means that the product of annihilators will annihilate a linear combination of functions. If \(L_1\) is the annihilator of \(y_1\) and \(L_2\) is the annihilator of \(y_2\): \(L_1 L_2 (y_1 + y_2) = L_1 L_2 y_1 + L_1 L_2 y_2\) Since \(L_1\) and \(L_2\) are annihilators, \(L_1 y_1 = 0\) and \(L_2 y_2 = 0\). So, \(L_1 L_2 y_1 = L_2 (L_1 y_1) = L_2 (0) = 0\). And \(L_1 L_2 y_2 = L_1 (L_2 y_2) = L_1 (0) = 0\). Therefore, \(L_1 L_2 (y_1 + y_2) = 0\).

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the Method of Annihilators.

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ y'' + 3y' + 2y = 0 \] Assume a solution of the form \(y = e^{mx}\). Then \(y' = me^{mx}\) and \(y'' = m^2e^{mx}\). Substitute these into the homogeneous equation: \[ m^2e^{mx} + 3me^{mx} + 2e^{mx} = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 + 3m + 2) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 + 3m + 2 = 0 \] Factor the quadratic equation: \[ (m+1)(m+2) = 0 \] This gives two distinct real roots: \[ m_1 = -1 \quad \text{and} \quad m_2 = -2 \] The complementary solution (homogeneous solution) is: \[ y_h = c_1 e^{-x} + c_2 e^{-2x} \]

Step 2: Find the Annihilator for \(f(x)\)

The non-homogeneous term is \(f(x) = 4x^2\). For a term of the form \(x^n\), the annihilator is \(D^{n+1}\). Here, \(f(x) = 4x^2\) (a polynomial of degree 2), so \(n=2\). The annihilator for \(4x^2\) is \(D^{2+1} = D^3\).

Step 3: Apply the Annihilator to the Original Equation

Apply the annihilator \(D^3\) to both sides of the original differential equation: \[ D^3(y'' + 3y' + 2y) = D^3(4x^2) \] Since \(D^3(4x^2) = 0\), we get: \[ D^3(D^2 + 3D + 2)y = 0 \] This is a new homogeneous differential equation.

Step 4: Solve the New Homogeneous Equation

The auxiliary equation for this new homogeneous equation is: \[ m^3(m^2 + 3m + 2) = 0 \] We already factored the quadratic part: \((m+1)(m+2)\). So, the auxiliary equation is: \[ m^3(m+1)(m+2) = 0 \] This equation has roots: \(m_1 = -1\) \(m_2 = -2\) \(m_3 = 0\) (with multiplicity 3)

The general solution to this new homogeneous equation is: \[ y(x) = C_1 e^{-x} + C_2 e^{-2x} + C_3 e^{0x} + C_4 x e^{0x} + C_5 x^2 e^{0x} \] \[ y(x) = C_1 e^{-x} + C_2 e^{-2x} + C_3 + C_4 x + C_5 x^2 \]

Step 5: Determine the Form of the Particular Solution (\(y_p\))

The terms in the general solution from Step 4 that are not part of \(y_h\) (from Step 1) form the particular solution \(y_p\). \(y_h = c_1 e^{-x} + c_2 e^{-2x}\) The new terms are \(C_3\), \(C_4 x\), and \(C_5 x^2\). So, the form of the particular solution is: \[ y_p(x) = Ax^2 + Bx + C \] (We use \(A, B, C\) instead of \(C_3, C_4, C_5\) for coefficients of \(y_p\)).

Step 6: Find the Derivatives of \(y_p\) and Substitute into the Original Equation

Calculate the derivatives of \(y_p\): \(y_p' = 2Ax + B\) \(y_p'' = 2A\)

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original non-homogeneous equation \(y'' + 3y' + 2y = 4x^2\): \[ (2A) + 3(2Ax + B) + 2(Ax^2 + Bx + C) = 4x^2 \] Expand and collect terms by powers of \(x\): \[ 2A + 6Ax + 3B + 2Ax^2 + 2Bx + 2C = 4x^2 \] \[ (2A)x^2 + (6A + 2B)x + (2A + 3B + 2C) = 4x^2 + 0x + 0 \]

Step 7: Equate Coefficients

Compare the coefficients of corresponding powers of \(x\) on both sides: Coefficient of \(x^2\): \(2A = 4 \implies A = 2\)

Coefficient of \(x\): \(6A + 2B = 0\) Substitute \(A=2\): \(6(2) + 2B = 0\) \(12 + 2B = 0\) \(2B = -12 \implies B = -6\)

Constant term: \(2A + 3B + 2C = 0\) Substitute \(A=2\) and \(B=-6\): \(2(2) + 3(-6) + 2C = 0\) \(4 - 18 + 2C = 0\) \(-14 + 2C = 0\) \(2C = 14 \implies C = 7\)

So, the particular solution is: \[ y_p(x) = 2x^2 - 6x + 7 \]

Step 8: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 e^{-x} + c_2 e^{-2x} + 2x^2 - 6x + 7 \]

Thus, the general solution to the differential equation is \(y = c_1 e^{-x} + c_2 e^{-2x} + 2x^2 - 6x + 7\).

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the Method of Undetermined Coefficients (which can be derived using the Annihilator Method, as shown below).

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ y'' - 3y' = 0 \] Assume a solution of the form \(y = e^{mx}\). Then \(y' = me^{mx}\) and \(y'' = m^2e^{mx}\). Substitute these into the homogeneous equation: \[ m^2e^{mx} - 3me^{mx} = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 - 3m) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 - 3m = 0 \] Factor the quadratic equation: \[ m(m-3) = 0 \] This gives two distinct real roots: \[ m_1 = 0 \quad \text{and} \quad m_2 = 3 \] The complementary solution (homogeneous solution) is: \[ y_h = c_1 e^{0x} + c_2 e^{3x} = c_1 + c_2 e^{3x} \]

Step 2: Find the Annihilators for \(f(x)\)

The non-homogeneous term is \(f(x) = 8e^{3x} + 4\sin x\). We consider each term separately.

For \(f_1(x) = 8e^{3x}\): The annihilator for \(e^{ax}\) is \((D-a)\). Here \(a=3\). So, the annihilator for \(8e^{3x}\) is \((D-3)\).

For \(f_2(x) = 4\sin x\): The annihilator for \(\sin(bx)\) or \(\cos(bx)\) is \((D^2 + b^2)\). Here \(b=1\). So, the annihilator for \(4\sin x\) is \((D^2 + 1^2) = (D^2 + 1)\).

The annihilator for the entire \(f(x)\) is the product of the individual annihilators: Annihilator \(= (D-3)(D^2+1)\).

Step 3: Apply the Annihilator to the Original Equation

The original differential equation can be written in operator form as \((D^2 - 3D)y = 8e^{3x} + 4\sin x\). Apply the annihilator \((D-3)(D^2+1)\) to both sides: \[ (D-3)(D^2+1)(D^2 - 3D)y = (D-3)(D^2+1)(8e^{3x} + 4\sin x) \] The right-hand side becomes 0 because \((D-3)\) annihilates \(e^{3x}\) and \((D^2+1)\) annihilates \(\sin x\). So, we get a new homogeneous differential equation: \[ D(D-3)(D^2+1)(D-3)y = 0 \] Rearranging and combining the \((D-3)\) terms: \[ D(D-3)^2(D^2+1)y = 0 \]

Step 4: Solve the New Homogeneous Equation

The auxiliary equation for this new homogeneous equation is: \[ m(m-3)^2(m^2+1) = 0 \] This equation has roots: \(m_1 = 0\) \(m_2 = 3\) (with multiplicity 2, due to the original factor and the annihilator factor) \(m_3 = \pm i\) (from \(m^2+1=0\))

The general solution to this new homogeneous equation is: \[ y(x) = C_1 e^{0x} + C_2 e^{3x} + C_3 x e^{3x} + C_4 \cos x + C_5 \sin x \] \[ y(x) = C_1 + C_2 e^{3x} + C_3 x e^{3x} + C_4 \cos x + C_5 \sin x \]

Step 5: Determine the Form of the Particular Solution (\(y_p\))

The terms in the general solution from Step 4 that are not part of \(y_h\) (from Step 1) form the particular solution \(y_p\). \(y_h = c_1 + c_2 e^{3x}\) The new terms are \(C_3 x e^{3x}\), \(C_4 \cos x\), and \(C_5 \sin x\). So, the form of the particular solution is: \[ y_p(x) = Axe^{3x} + B\cos x + C\sin x \] (We use \(A, B, C\) for coefficients of \(y_p\)).

Step 6: Find the Derivatives of \(y_p\) and Substitute into the Original Equation

Calculate the derivatives of \(y_p\): \(y_p = Axe^{3x} + B\cos x + C\sin x\)

\(y_p' = A(e^{3x} + 3xe^{3x}) - B\sin x + C\cos x = A e^{3x}(1+3x) - B\sin x + C\cos x\)

\(y_p'' = A(3e^{3x}(1+3x) + e^{3x}(3)) - B\cos x - C\sin x\) \(y_p'' = A(3e^{3x} + 9xe^{3x} + 3e^{3x}) - B\cos x - C\sin x\) \(y_p'' = A(6e^{3x} + 9xe^{3x}) - B\cos x - C\sin x = Ae^{3x}(6+9x) - B\cos x - C\sin x\)

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original non-homogeneous equation \(y'' - 3y' = 8e^{3x} + 4\sin x\): \[ [Ae^{3x}(6+9x) - B\cos x - C\sin x] - 3[A e^{3x}(1+3x) - B\sin x + C\cos x] = 8e^{3x} + 4\sin x \] Expand and collect terms: \[ 6Ae^{3x} + 9Axe^{3x} - B\cos x - C\sin x - 3Ae^{3x} - 9Axe^{3x} + 3B\sin x - 3C\cos x = 8e^{3x} + 4\sin x \] Combine terms: \((6A - 3A)e^{3x} + (9A - 9A)xe^{3x} + (-B - 3C)\cos x + (-C + 3B)\sin x = 8e^{3x} + 4\sin x\) \[ 3Ae^{3x} + (3B - C)\sin x + (-B - 3C)\cos x = 8e^{3x} + 4\sin x \]

Step 7: Equate Coefficients

Compare the coefficients of corresponding functions on both sides:

Coefficient of \(e^{3x}\): \(3A = 8 \implies A = \frac{8}{3}\)

Coefficient of \(\sin x\): \(3B - C = 4\) (Equation 1)

Coefficient of \(\cos x\): \(-B - 3C = 0 \implies B = -3C\) (Equation 2)

Substitute Equation 2 into Equation 1: \(3(-3C) - C = 4\) \(-9C - C = 4\) \(-10C = 4 \implies C = -\frac{4}{10} = -\frac{2}{5}\)

Now find \(B\) using Equation 2: \(B = -3C = -3\left(-\frac{2}{5}\right) = \frac{6}{5}\)

So, the particular solution is: \[ y_p(x) = \frac{8}{3}xe^{3x} + \frac{6}{5}\cos x - \frac{2}{5}\sin x \]

Step 8: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 + c_2 e^{3x} + \frac{8}{3}xe^{3x} + \frac{6}{5}\cos x - \frac{2}{5}\sin x \]

Thus, the general solution to the differential equation is \(y = c_1 + c_2 e^{3x} + \frac{8}{3}xe^{3x} + \frac{6}{5}\cos x - \frac{2}{5}\sin x\).

Superposition Approach (Trial Solution Method):

“Assume a particular solution \(y_p\) based on the form of \(g(x)\).”

  • For Polynomials: If \(g(x)\) is a polynomial of degree \(n\), assume \(y_p\) is a general polynomial of degree \(n\): \(A_n x^n + \dots + A_1 x + A_0\).
  • For Trigonometry: If \(g(x)\) contains \(\sin(kx)\) or \(\cos(kx)\), assume \(y_p = A \cos(kx) + B \sin(kx)\).
  • For Exponentials: If \(g(x)\) contains \(e^{ax}\), assume \(y_p = A e^{ax}\).
  • For Sums/Products:
    • If \(g(x)\) is a sum of two or more terms (e.g., \(g_1(x) + g_2(x)\)), find \(y_{p_1}\) for \(g_1(x)\) and \(y_{p_2}\) for \(g_2(x)\) separately, then \(y_p = y_{p_1} + y_{p_2}\).
    • If \(g(x)\) is a product (e.g., \(x^k e^{ax} \cos(\beta x)\)), assume \(y_p\) is a product of the assumed forms for each factor.

Important Rule: - If a term in the assumed \(y_p\) is already part of the complementary solution \(y_c\), multiply the assumed \(y_p\) term by the lowest power of \(x\) (usually \(x\) or \(x^2\)) that eliminates the duplication. This is called the multiplication rule. - For example, if \(g(x) = C e^{ax}\) and \(e^{ax}\) is already in \(y_c\), assume \(y_p = Ax e^{ax}\). If \(x e^{ax}\) is also in \(y_c\), assume \(y_p = Ax^2 e^{ax}\).

Table of Suggested Particular Solutions \(Y_p\):

\(g(x)\) (Term) Form of \(Y_p\)
\(c\) (constant) \(A\)
\(c x^n\) \(A_n x^n + \dots + A_1 x + A_0\)
\(c e^{ax}\) \(A e^{ax}\)
\(c \sin(kx)\) or \(c \cos(kx)\) \(A \cos(kx) + B \sin(kx)\)
\(c x^n e^{ax}\) \((A_n x^n + \dots + A_0) e^{ax}\)
\(c x^n \cos(kx)\) or \(c x^n \sin(kx)\) \((A_n x^n + \dots + A_0) \cos(kx) + (B_n x^n + \dots + B_0) \sin(kx)\)
\(c e^{ax} \cos(kx)\) or \(c e^{ax} \sin(kx)\) \(e^{ax}(A \cos(kx) + B \sin(kx))\)
\(c x^n e^{ax} \cos(kx)\) or \(c x^n e^{ax} \sin(kx)\) \(e^{ax}((A_n x^n + \dots + A_0) \cos(kx) + (B_n x^n + \dots + B_0) \sin(kx))\)

Steps to Solve Non-Homogeneous Equation using Undetermined Coefficients:

  1. Solve the associated homogeneous equation \(Ly = 0\) to find the complementary solution \(y_c\).
  2. Determine the form of the particular solution \(y_p\) based on \(g(x)\) and applying the multiplication rule if necessary (i.e., if terms in \(y_p\) duplicate terms in \(y_c\)).
  3. Calculate the derivatives of \(y_p\) (up to the order of the differential equation).
  4. Substitute \(y_p\) and its derivatives into the original non-homogeneous equation \(Ly = g(x)\).
  5. Equate the coefficients of like terms on both sides of the equation to form a system of linear equations for the undetermined coefficients.
  6. Solve the system of equations to find the values of the coefficients.
  7. Write the general solution as \(y = y_c + y_p\).

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the Method of Undetermined Coefficients (Superposition Method).

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ y'' + y = 0 \] Assume a solution of the form \(y = e^{mx}\). Substituting this into the homogeneous equation gives the auxiliary equation: \[ m^2 + 1 = 0 \] Solving for \(m\): \[ m^2 = -1 \implies m = \pm i \] The roots are complex conjugates \(m = \alpha \pm i\beta\), where \(\alpha = 0\) and \(\beta = 1\). The complementary solution (homogeneous solution) is: \[ y_h = e^{0x}(c_1 \cos(1x) + c_2 \sin(1x)) = c_1 \cos x + c_2 \sin x \]

Step 2: Determine the Form of the Particular Solution (\(y_p\)) using Superposition

The non-homogeneous term is \(f(x) = x \cos x - \cos x = (x-1)\cos x\). Since the right-hand side is a product of a polynomial and a cosine function, and the homogeneous solution contains \(\cos x\) and \(\sin x\) (which are part of the family of functions generated by \((x-1)\cos x\)), we need to multiply the standard form of \(y_p\) by \(x\).

The general form for \(P_n(x) \cos(\beta x)\) or \(P_n(x) \sin(\beta x)\) where \(\beta\) is a root of the auxiliary equation is \(x^s [ (A_n x^n + \dots + A_0)\cos(\beta x) + (B_n x^n + \dots + B_0)\sin(\beta x) ]\). Here, \(P(x) = x-1\) is a polynomial of degree \(n=1\). The value \(\beta = 1\) is a root of the auxiliary equation with multiplicity \(s=1\) (since \(m=\pm i\) are roots).

Therefore, the form of the particular solution is: \[ y_p(x) = x [ (Ax+B)\cos x + (Cx+D)\sin x ] \] \[ y_p(x) = (Ax^2+Bx)\cos x + (Cx^2+Dx)\sin x \]

Step 3: Find the Derivatives of \(y_p\) and Substitute into the Original Equation

Calculate the derivatives of \(y_p\): \(y_p = (Ax^2+Bx)\cos x + (Cx^2+Dx)\sin x\)

\(y_p' = [(2Ax+B)\cos x - (Ax^2+Bx)\sin x] + [(2Cx+D)\sin x + (Cx^2+Dx)\cos x]\) \(y_p' = [(2Ax+B) + (Cx^2+Dx)]\cos x + [(2Cx+D) - (Ax^2+Bx)]\sin x\)

\(y_p'' = \frac{d}{dx}\{ [(2Ax+B) + (Cx^2+Dx)]\cos x \} + \frac{d}{dx}\{ [(2Cx+D) - (Ax^2+Bx)]\sin x \}\) \(y_p'' = [ (2A+2Cx+D)\cos x - (2Ax+B+Cx^2+Dx)\sin x ] + [ (2C-2Ax-B)\sin x + (2Cx+D-Ax^2-Bx)\cos x ]\)

Group terms for \(\cos x\) and \(\sin x\) in \(y_p''\): Coefficient of \(\cos x\): \((2A+2Cx+D) + (2Cx+D-Ax^2-Bx) = (-Ax^2-Bx) + (2A+4Cx+2D)\) Coefficient of \(\sin x\): \(-(2Ax+B+Cx^2+Dx) + (2C-2Ax-B) = (-Cx^2-Dx) + (-4Ax-2B+2C)\)

Now, substitute \(y_p\) and \(y_p''\) into the original non-homogeneous equation \(y'' + y = (x-1)\cos x\): $y_p’’ + y_p = $ \((-Ax^2-Bx + 2A+4Cx+2D)\cos x + (-Cx^2-Dx - 4Ax-2B+2C)\sin x\) \(+ (Ax^2+Bx)\cos x + (Cx^2+Dx)\sin x\) \(= (2A+4Cx+2D)\cos x + (-4Ax-2B+2C)\sin x\)

Equate this to \((x-1)\cos x + 0 \sin x\): \[ (2A+4Cx+2D)\cos x + (-4Ax-2B+2C)\sin x = (x-1)\cos x \]

Step 4: Equate Coefficients

Compare the coefficients of corresponding functions and powers of \(x\) on both sides:

Coefficients of \(\cos x\): \(2A + 4Cx + 2D = x - 1\) Comparing the coefficient of \(x\): \(4C = 1 \implies C = \frac{1}{4}\) Comparing the constant term: \(2A + 2D = -1\) (Equation 1)

Coefficients of \(\sin x\): \(-4Ax - 2B + 2C = 0\) Comparing the coefficient of \(x\): \(-4A = 0 \implies A = 0\) Comparing the constant term: \(-2B + 2C = 0 \implies -2B + 2(\frac{1}{4}) = 0 \implies -2B + \frac{1}{2} = 0 \implies 2B = \frac{1}{2} \implies B = \frac{1}{4}\)

Now use \(A=0\) in Equation 1: \(2(0) + 2D = -1 \implies 2D = -1 \implies D = -\frac{1}{2}\)

So, the coefficients are: \(A = 0\) \(B = \frac{1}{4}\) \(C = \frac{1}{4}\) \(D = -\frac{1}{2}\)

The particular solution is: \[ y_p(x) = (0 \cdot x^2 + \frac{1}{4}x)\cos x + (\frac{1}{4}x^2 - \frac{1}{2}x)\sin x \] \[ y_p(x) = \frac{1}{4}x\cos x + (\frac{1}{4}x^2 - \frac{1}{2}x)\sin x \]

Step 5: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 \cos x + c_2 \sin x + \frac{1}{4}x\cos x + (\frac{1}{4}x^2 - \frac{1}{2}x)\sin x \] We can also write it as: \[ y(x) = c_1 \cos x + c_2 \sin x + \frac{1}{4}x\cos x + \frac{1}{4}x^2\sin x - \frac{1}{2}x\sin x \]

Thus, the general solution to the differential equation is \(y = c_1 \cos x + c_2 \sin x + \frac{1}{4}x\cos x + (\frac{1}{4}x^2 - \frac{1}{2}x)\sin x\).

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the Method of Undetermined Coefficients (Superposition Method).

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ y'' - 2y' + y = 0 \] Assume a solution of the form \(y = e^{mx}\). Substituting this into the homogeneous equation gives the auxiliary equation: \[ m^2 - 2m + 1 = 0 \] Factor the quadratic equation: \[ (m-1)^2 = 0 \] This gives a repeated real root: \[ m = 1 \] The complementary solution (homogeneous solution) is: \[ y_h = c_1 e^{x} + c_2 x e^{x} \]

Step 2: Determine the Form of the Particular Solution (\(y_p\))

The non-homogeneous term is \(f(x) = 10e^{-2x}\cos x\). This is of the form \(e^{ax}P_n(x)\cos(bx)\), where \(a=-2\), \(P_n(x)=10\) (a polynomial of degree \(n=0\)), and \(b=1\). The standard form for \(y_p\) for such a term is \(e^{ax}[(A_n x^n + \dots + A_0)\cos(bx) + (B_n x^n + \dots + B_0)\sin(bx)]\). In our case, \(a=-2\), \(b=1\), and \(n=0\). So, the initial guess for \(y_p\) is \(e^{-2x}[A\cos x + B\sin x]\). Now, we check for duplication with terms in \(y_h\). The terms in \(y_h\) are \(e^x\) and \(xe^x\). The exponential factor in \(y_p\) is \(e^{-2x}\), which is different from \(e^x\). Therefore, there is no duplication, and we do not need to multiply by \(x^s\).

The form of the particular solution is: \[ y_p(x) = Ae^{-2x}\cos x + Be^{-2x}\sin x \]

Step 3: Find the Derivatives of \(y_p\) and Substitute into the Original Equation

Calculate the derivatives of \(y_p\): \(y_p = Ae^{-2x}\cos x + Be^{-2x}\sin x\)

\(y_p' = [ -2Ae^{-2x}\cos x - Ae^{-2x}\sin x ] + [ -2Be^{-2x}\sin x + Be^{-2x}\cos x ]\) \(y_p' = e^{-2x}[ (-2A+B)\cos x + (-A-2B)\sin x ]\)

\(y_p'' = -2e^{-2x}[ (-2A+B)\cos x + (-A-2B)\sin x ] + e^{-2x}[ (-2A+B)(-\sin x) + (-A-2B)\cos x ]\) \(y_p'' = e^{-2x}[ (4A-2B)\cos x + (2A+4B)\sin x - (-2A+B)\sin x + (-A-2B)\cos x ]\) \(y_p'' = e^{-2x}[ (4A-2B-A-2B)\cos x + (2A+4B+2A-B)\sin x ]\) \(y_p'' = e^{-2x}[ (3A-4B)\cos x + (4A+3B)\sin x ]\)

Substitute \(y_p\), \(y_p'\), and \(y_p''\) into the original non-homogeneous equation \(y'' - 2y' + y = 10e^{-2x}\cos x\): \(e^{-2x}[ (3A-4B)\cos x + (4A+3B)\sin x ] - 2e^{-2x}[ (-2A+B)\cos x + (-A-2B)\sin x ] + e^{-2x}[ A\cos x + B\sin x ] = 10e^{-2x}\cos x\)

Divide by \(e^{-2x}\) (since \(e^{-2x} \neq 0\)): \((3A-4B)\cos x + (4A+3B)\sin x - 2(-2A+B)\cos x - 2(-A-2B)\sin x + A\cos x + B\sin x = 10\cos x\)

Combine coefficients of \(\cos x\): \((3A-4B) - 2(-2A+B) + A = 3A-4B+4A-2B+A = (3+4+1)A + (-4-2)B = 8A-6B\)

Combine coefficients of \(\sin x\): \((4A+3B) - 2(-A-2B) + B = 4A+3B+2A+4B+B = (4+2)A + (3+4+1)B = 6A+8B\)

So the equation becomes: \[ (8A-6B)\cos x + (6A+8B)\sin x = 10\cos x \]

Step 4: Equate Coefficients

Compare the coefficients of corresponding functions on both sides:

Coefficient of \(\cos x\): \(8A - 6B = 10\) (Equation 1)

Coefficient of \(\sin x\): \(6A + 8B = 0\) (Equation 2)

From Equation 2, we can express \(A\) in terms of \(B\): \(6A = -8B \implies A = -\frac{8}{6}B = -\frac{4}{3}B\)

Substitute this into Equation 1: \(8(-\frac{4}{3}B) - 6B = 10\) \(-\frac{32}{3}B - 6B = 10\) To combine the \(B\) terms, find a common denominator: \(-\frac{32}{3}B - \frac{18}{3}B = 10\) \(-\frac{50}{3}B = 10\) \(B = 10 \cdot \left(-\frac{3}{50}\right) = -\frac{30}{50} = -\frac{3}{5}\)

Now find \(A\): \(A = -\frac{4}{3}B = -\frac{4}{3}\left(-\frac{3}{5}\right) = \frac{12}{15} = \frac{4}{5}\)

So, the coefficients are: \(A = \frac{4}{5}\) \(B = -\frac{3}{5}\)

The particular solution is: \[ y_p(x) = \frac{4}{5}e^{-2x}\cos x - \frac{3}{5}e^{-2x}\sin x \]

Step 5: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 e^{x} + c_2 x e^{x} + \frac{4}{5}e^{-2x}\cos x - \frac{3}{5}e^{-2x}\sin x \]

Thus, the general solution to the differential equation is \(y = c_1 e^{x} + c_2 x e^{x} + \frac{4}{5}e^{-2x}\cos x - \frac{3}{5}e^{-2x}\sin x\).

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Variation of Parameters

This is a general method to find a particular solution \(y_p\) for any non-homogeneous linear differential equation, even if the coefficients are not constant or \(g(x)\) is not one of the types suitable for Undetermined Coefficients.

Given the non-homogeneous second-order differential equation: \(a_2(x)y'' + a_1(x)y' + a_0(x)y = g(x)\)

  1. Convert to Standard Form: Divide by \(a_2(x)\) to get the equation in standard form: \(y'' + P(x)y' + Q(x)y = f(x)\), where \(f(x) = g(x)/a_2(x)\).
  2. Find the Complementary Solution: Solve the associated homogeneous equation \(y'' + P(x)y' + Q(x)y = 0\) to find the complementary solution \(y_c = c_1 y_1(x) + c_2 y_2(x)\).
  3. Calculate the Wronskian \(W(y_1, y_2)\): \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1' \] Note: \(W\) must be non-zero for \(y_1, y_2\) to be a fundamental set of solutions.
  4. Calculate \(W_1\) and \(W_2\): \[ W_1 = \begin{vmatrix} 0 & y_2 \\ f(x) & y_2' \end{vmatrix} = -y_2 f(x) \] \[ W_2 = \begin{vmatrix} y_1 & 0 \\ y_1' & f(x) \end{vmatrix} = y_1 f(x) \]
  5. Find \(u_1'\) and \(u_2'\): Assume the particular solution is of the form \(y_p = u_1(x)y_1(x) + u_2(x)y_2(x)\). The derivatives \(u_1'\) and \(u_2'\) are given by: \[ u_1' = \frac{W_1}{W} = \frac{-y_2 f(x)}{W(y_1, y_2)} \] \[ u_2' = \frac{W_2}{W} = \frac{y_1 f(x)}{W(y_1, y_2)} \]
  6. Integrate to find \(u_1\) and \(u_2\): \(u_1(x) = \int u_1'(x)\,dx\) \(u_2(x) = \int u_2'(x)\,dx\) Important: Do NOT introduce constants of integration here. These constants would simply replicate terms already present in \(y_c\).
  7. Form the Particular Solution: \(y_p = u_1(x)y_1(x) + u_2(x)y_2(x)\)
  8. Write the General Solution: \(y = y_c + y_p\)

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the method of Variation of Parameters.

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ y'' - 4y' + 4y = 0 \] Assume a solution of the form \(y = e^{mx}\). Then \(y' = me^{mx}\) and \(y'' = m^2e^{mx}\). Substitute these into the homogeneous equation: \[ m^2e^{mx} - 4me^{mx} + 4e^{mx} = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 - 4m + 4) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 - 4m + 4 = 0 \] Factor the quadratic equation: \[ (m-2)^2 = 0 \] This gives a repeated real root: \[ m = 2 \] For a repeated real root \(m\), the complementary solution (homogeneous solution) is: \[ y_h = c_1 e^{mx} + c_2 x e^{mx} \] Substituting \(m=2\): \[ y_h = c_1 e^{2x} + c_2 x e^{2x} \] From this, we identify two linearly independent solutions for the homogeneous equation: \(y_1(x) = e^{2x}\) \(y_2(x) = x e^{2x}\)

Step 2: Identify \(f(x)\)

The differential equation is already in the standard form \(y'' + P(x)y' + Q(x)y = f(x)\). From the given equation, \(f(x) = (x+1)e^{2x}\).

Step 3: Calculate the Wronskian \(W(y_1, y_2)\)

The Wronskian of \(y_1\) and \(y_2\) is given by: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1' \] We have \(y_1 = e^{2x}\) and \(y_2 = x e^{2x}\). \(y_1' = 2e^{2x}\) \(y_2' = \frac{d}{dx}(x e^{2x}) = 1 \cdot e^{2x} + x \cdot (2e^{2x}) = e^{2x} + 2x e^{2x} = e^{2x}(1+2x)\) \[ W(e^{2x}, x e^{2x}) = e^{2x} (e^{2x}(1+2x)) - (x e^{2x}) (2e^{2x}) \] \[ = e^{4x}(1+2x) - 2x e^{4x} \] \[ = e^{4x} + 2x e^{4x} - 2x e^{4x} \] \[ W = e^{4x} \]

Step 4: Calculate \(W_1\) and \(W_2\)

\(W_1\) is obtained by replacing the first column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_1 = \begin{vmatrix} 0 & y_2 \\ f(x) & y_2' \end{vmatrix} = 0 \cdot y_2' - y_2 \cdot f(x) = -y_2 f(x) \] \[ W_1 = -(x e^{2x}) ((x+1)e^{2x}) = -x(x+1)e^{4x} = -(x^2+x)e^{4x} \]

\(W_2\) is obtained by replacing the second column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_2 = \begin{vmatrix} y_1 & 0 \\ y_1' & f(x) \end{vmatrix} = y_1 \cdot f(x) - 0 \cdot y_1' = y_1 f(x) \] \[ W_2 = e^{2x} ((x+1)e^{2x}) = (x+1)e^{4x} \]

Step 5: Find \(u_1'\) and \(u_2'\)

The derivatives of the functions \(u_1(x)\) and \(u_2(x)\) are given by: \[ u_1' = \frac{W_1}{W} = \frac{-(x^2+x)e^{4x}}{e^{4x}} = -(x^2+x) = -x^2 - x \] \[ u_2' = \frac{W_2}{W} = \frac{(x+1)e^{4x}}{e^{4x}} = x+1 \]

Step 6: Integrate \(u_1'\) and \(u_2'\) to Find \(u_1\) and \(u_2\)

Integrate \(u_1'\): \[ u_1 = \int (-x^2 - x) \, dx = -\frac{x^3}{3} - \frac{x^2}{2} \]

Integrate \(u_2'\): \[ u_2 = \int (x+1) \, dx = \frac{x^2}{2} + x \]

Step 7: Form the Particular Solution \(y_p\)

The particular solution is given by \(y_p = u_1 y_1 + u_2 y_2\): \[ y_p = \left(-\frac{x^3}{3} - \frac{x^2}{2}\right) e^{2x} + \left(\frac{x^2}{2} + x\right) x e^{2x} \] \[ y_p = -\frac{x^3}{3}e^{2x} - \frac{x^2}{2}e^{2x} + \frac{x^3}{2}e^{2x} + x^2e^{2x} \] Combine like terms (terms with \(x^3e^{2x}\) and \(x^2e^{2x}\)): For \(x^3e^{2x}\) terms: \(-\frac{1}{3} + \frac{1}{2} = -\frac{2}{6} + \frac{3}{6} = \frac{1}{6}\) For \(x^2e^{2x}\) terms: \(-\frac{1}{2} + 1 = \frac{1}{2}\) \[ y_p = \frac{1}{6}x^3 e^{2x} + \frac{1}{2}x^2 e^{2x} \]

Step 8: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 e^{2x} + c_2 x e^{2x} + \frac{1}{6}x^3 e^{2x} + \frac{1}{2}x^2 e^{2x} \]

Thus, the general solution to the differential equation is \(y = c_1 e^{2x} + c_2 x e^{2x} + \frac{1}{6}x^3 e^{2x} + \frac{1}{2}x^2 e^{2x}\).

This is a non-homogeneous linear second-order differential equation with constant coefficients. We will solve it using the method of Variation of Parameters.

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ 4y'' + 36y = 0 \] Divide by 4 to get the standard form: \[ y'' + 9y = 0 \] Assume a solution of the form \(y = e^{mx}\). Then \(y' = me^{mx}\) and \(y'' = m^2e^{mx}\). Substitute these into the homogeneous equation: \[ m^2e^{mx} + 9e^{mx} = 0 \] Factor out \(e^{mx}\) (since \(e^{mx} \neq 0\)): \[ e^{mx}(m^2 + 9) = 0 \] The auxiliary (or characteristic) equation is: \[ m^2 + 9 = 0 \] Solve for \(m\): \[ m^2 = -9 \] \[ m = \pm \sqrt{-9} = \pm 3i \] The roots are complex conjugates of the form \(m = \alpha \pm i\beta\), where \(\alpha = 0\) and \(\beta = 3\). The complementary solution (homogeneous solution) is: \[ y_h = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x)) \] Substituting \(\alpha = 0\) and \(\beta = 3\): \[ y_h = e^{0x}(c_1 \cos(3x) + c_2 \sin(3x)) \] \[ y_h = c_1 \cos(3x) + c_2 \sin(3x) \] From this, we identify two linearly independent solutions for the homogeneous equation: \(y_1(x) = \cos(3x)\) \(y_2(x) = \sin(3x)\)

Step 2: Identify \(f(x)\)

The method of Variation of Parameters requires the differential equation to be in the standard form \(y'' + P(x)y' + Q(x)y = f(x)\). Divide the original non-homogeneous equation by 4: \[ y'' + 9y = \frac{1}{4}\csc(3x) \] From this, we identify \(f(x) = \frac{1}{4}\csc(3x)\).

Step 3: Calculate the Wronskian \(W(y_1, y_2)\)

The Wronskian of \(y_1\) and \(y_2\) is given by: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1' \] We have \(y_1 = \cos(3x)\) and \(y_2 = \sin(3x)\). \(y_1' = -3\sin(3x)\) \(y_2' = 3\cos(3x)\) \[ W(\cos(3x), \sin(3x)) = \cos(3x)(3\cos(3x)) - \sin(3x)(-3\sin(3x)) \] \[ = 3\cos^2(3x) + 3\sin^2(3x) \] \[ = 3(\cos^2(3x) + \sin^2(3x)) \] Since \(\cos^2(\theta) + \sin^2(\theta) = 1\): \[ W = 3 \]

Step 4: Calculate \(W_1\) and \(W_2\)

\(W_1\) is obtained by replacing the first column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_1 = \begin{vmatrix} 0 & y_2 \\ f(x) & y_2' \end{vmatrix} = 0 \cdot y_2' - y_2 \cdot f(x) = -y_2 f(x) \] \[ W_1 = -\sin(3x) \left( \frac{1}{4}\csc(3x) \right) \] Since \(\csc(3x) = \frac{1}{\sin(3x)}\): \[ W_1 = -\sin(3x) \left( \frac{1}{4\sin(3x)} \right) = -\frac{1}{4} \]

\(W_2\) is obtained by replacing the second column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_2 = \begin{vmatrix} y_1 & 0 \\ y_1' & f(x) \end{vmatrix} = y_1 \cdot f(x) - 0 \cdot y_1' = y_1 f(x) \] \[ W_2 = \cos(3x) \left( \frac{1}{4}\csc(3x) \right) \] \[ W_2 = \frac{1}{4} \frac{\cos(3x)}{\sin(3x)} = \frac{1}{4} \cot(3x) \]

Step 5: Find \(u_1'\) and \(u_2'\)

The derivatives of the functions \(u_1(x)\) and \(u_2(x)\) are given by: \[ u_1' = \frac{W_1}{W} = \frac{-1/4}{3} = -\frac{1}{12} \] \[ u_2' = \frac{W_2}{W} = \frac{\frac{1}{4}\cot(3x)}{3} = \frac{1}{12}\cot(3x) \]

Step 6: Integrate \(u_1'\) and \(u_2'\) to Find \(u_1\) and \(u_2\)

Integrate \(u_1'\): \[ u_1 = \int -\frac{1}{12} \, dx = -\frac{1}{12} x \]

Integrate \(u_2'\): \[ u_2 = \int \frac{1}{12}\cot(3x) \, dx \] Recall that \(\int \cot(ax) \, dx = \frac{1}{a} \ln|\sin(ax)|\). \[ u_2 = \frac{1}{12} \cdot \frac{1}{3} \ln|\sin(3x)| = \frac{1}{36} \ln|\sin(3x)| \]

Step 7: Form the Particular Solution \(y_p\)

The particular solution is given by \(y_p = u_1 y_1 + u_2 y_2\): \[ y_p = \left(-\frac{1}{12} x\right) \cos(3x) + \left(\frac{1}{36} \ln|\sin(3x)|\right) \sin(3x) \] \[ y_p = -\frac{x}{12}\cos(3x) + \frac{1}{36}\sin(3x)\ln|\sin(3x)| \]

Step 8: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 \cos(3x) + c_2 \sin(3x) - \frac{x}{12}\cos(3x) + \frac{1}{36}\sin(3x)\ln|\sin(3x)| \]

Thus, the general solution to the differential equation is \(y = c_1 \cos(3x) + c_2 \sin(3x) - \frac{x}{12}\cos(3x) + \frac{1}{36}\sin(3x)\ln|\sin(3x)|\).

For \(n^\text{th}\) order differential equations: The particular solution will be \(y_p = u_1 y_1 + u_2 y_2 + \dots + u_n y_n\). The Wronskian \(W\) will be an \(n \times n\) determinant of the fundamental solutions and their derivatives. To find \(u_k'\), replace the \(k\)-th column of the Wronskian matrix with \((0, 0, \ldots, 0, f(x))^T\) (where \(f(x)\) is the right-hand side in standard form, and the other elements in the last row are zeros), then divide by \(W\).

Why we don’t introduce constants of integration with \(u_1, u_2\):

If we included constants of integration, say \(u_1 = \int u_1' dx + C_1\) and \(u_2 = \int u_2' dx + C_2\), then: \(y_p = (\int u_1' dx + C_1)y_1 + (\int u_2' dx + C_2)y_2\) \(y_p = (\int u_1' dx)y_1 + (\int u_2' dx)y_2 + C_1 y_1 + C_2 y_2\) The terms \(C_1 y_1 + C_2 y_2\) are already part of the complementary solution \(y_c\). Since the general solution is \(y = y_c + y_p\), these extra terms are redundant and would simply be absorbed into the arbitrary constants of \(y_c\).

Cauchy-Euler Equation

General form: \(a_n x^n \frac{d^n y}{dx^n} + a_{n-1} x^{n-1} \frac{d^{n-1} y}{dx^{n-1}} + \cdots + a_1 x \frac{dy}{dx} + a_0 y = g(x)\)

This is a linear differential equation with variable coefficients, specifically, the power of \(x\) matches the order of the derivative.

Method of Solution (Homogeneous Case):

For the homogeneous second-order Cauchy-Euler equation: \(ax^2 \frac{d^2 y}{dx^2} + bx \frac{dy}{dx} + cy = 0\)

  1. Assume a solution of the form \(y = x^m\).
  2. Find the derivatives: \(y' = m x^{m-1}\) \(y'' = m(m-1) x^{m-2}\)
  3. Substitute into the differential equation: \(a x^2 [m(m-1) x^{m-2}] + b x [m x^{m-1}] + c x^m = 0\) \(a m(m-1) x^m + b m x^m + c x^m = 0\)
  4. Factor out \(x^m\): \(x^m [a m(m-1) + b m + c] = 0\)
  5. Form the Auxiliary (Indicial) Equation: Since \(x^m \ne 0\) (for the relevant interval \(x > 0\)), we must have: \(a m(m-1) + b m + c = 0\) \(am^2 - am + bm + c = 0\) \(am^2 + (b-a)m + c = 0\) This is a quadratic equation in \(m\). Solve for \(m\) to find the roots \(m_1\) and \(m_2\).

Types of Roots and Corresponding Solutions (for Second Order):

  1. Real and Distinct Roots (\(m_1 \ne m_2\)): The general solution is \(y = c_1 x^{m_1} + c_2 x^{m_2}\).

  2. Real and Equal Roots (\(m_1 = m_2 = m\)): The general solution is \(y = c_1 x^m + c_2 x^m \ln|x|\).

    Derivation of second solution using Reduction of Order: Given \(y_1 = x^m\), and for the standard form \(y'' + P(x)y' + Q(x)y = 0\): \(P(x) = \frac{bx}{ax^2} = \frac{b}{ax}\). We use the shortcut formula: \[ y_2 = y_1(x) \int \frac{e^{-\int P(x)\,dx}}{[y_1(x)]^2}\,dx \] First, calculate \(-\int P(x)\,dx = -\int \frac{b}{ax}\,dx = -\frac{b}{a} \ln|x| = \ln(|x|^{-b/a})\). Then, \(e^{-\int P(x)\,dx} = e^{\ln(|x|^{-b/a})} = |x|^{-b/a}\). So, \[ y_2 = x^m \int \frac{|x|^{-b/a}}{(x^m)^2}\,dx = x^m \int \frac{x^{-b/a}}{x^{2m}}\,dx = x^m \int x^{-b/a - 2m}\,dx \] Since \(m = -\frac{(b-a)}{2a}\) for repeated roots, we have \(2m = -\frac{b-a}{a} = \frac{a-b}{a} = 1 - \frac{b}{a}\). So, \(-b/a - 2m = -b/a - (1 - b/a) = -b/a - 1 + b/a = -1\). Therefore, \[ y_2 = x^m \int x^{-1}\,dx = x^m \int \frac{1}{x}\,dx = x^m \ln|x| \] Thus, the general solution for repeated roots is \(y = c_1 x^m + c_2 x^m \ln|x|\).

  3. Complex Conjugate Roots (\(m_1 = \alpha + i\beta\), \(m_2 = \alpha - i\beta\)): Using \(x^m = x^{\alpha + i\beta} = x^\alpha x^{i\beta} = x^\alpha e^{i\beta \ln|x|}\) (since \(x^A = e^{A \ln x}\)) Applying Euler’s Formula, \(e^{i\theta} = \cos\theta + i\sin\theta\): \(x^{i\beta} = \cos(\beta \ln|x|) + i\sin(\beta \ln|x|)\) So, \(x^m = x^\alpha [\cos(\beta \ln|x|) + i\sin(\beta \ln|x|)]\). The two linearly independent real solutions are obtained from the real and imaginary parts: \(y_1 = x^\alpha \cos(\beta \ln|x|)\) \(y_2 = x^\alpha \sin(\beta \ln|x|)\) The general solution is \(y = x^\alpha [c_1 \cos(\beta \ln|x|) + c_2 \sin(\beta \ln|x|)]\).

This is a homogeneous linear second-order Cauchy-Euler (or Euler-Cauchy) differential equation, characterized by terms of the form \(x^k y^{(k)}\).

Step 1: Assume a Solution Form

For Cauchy-Euler equations, we assume a solution of the form \(y = x^m\). Then we find its derivatives: \(y' = mx^{m-1}\) \(y'' = m(m-1)x^{m-2}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ x^2 [m(m-1)x^{m-2}] - 3x [mx^{m-1}] + 4 [x^m] = 0 \] \[ m(m-1)x^{m-2}x^2 - 3mx^{m-1}x^1 + 4x^m = 0 \] \[ m(m-1)x^m - 3mx^m + 4x^m = 0 \] Factor out \(x^m\) (since \(x > 0\), \(x^m \neq 0\)): \[ x^m [m(m-1) - 3m + 4] = 0 \] Since \(x^m \neq 0\), the auxiliary (or characteristic) equation is: \[ m(m-1) - 3m + 4 = 0 \]

Step 3: Solve the Auxiliary Equation

Expand and simplify the auxiliary equation: \[ m^2 - m - 3m + 4 = 0 \] \[ m^2 - 4m + 4 = 0 \] This is a quadratic equation. We can solve it by factoring: \[ (m-2)(m-2) = 0 \] \[ (m-2)^2 = 0 \] This gives a repeated root: \[ m = 2 \]

Step 4: Write the General Solution based on the Roots

For a Cauchy-Euler equation with a repeated root \(m_1 = m_2 = m\), the general solution is of the form: \[ y = c_1 x^m + c_2 x^m \ln|x| \] Since \(x > 0\) is given, we can write \(\ln x\). Substitute \(m=2\): \[ y = c_1 x^2 + c_2 x^2 \ln x \] where \(c_1\) and \(c_2\) are arbitrary constants.

Thus, the general solution to the differential equation is \(y = c_1 x^2 + c_2 x^2 \ln x\).

This is a homogeneous linear second-order Cauchy-Euler (or Euler-Cauchy) differential equation.

Step 1: Assume a Solution Form

For Cauchy-Euler equations, we assume a solution of the form \(y = x^m\). Then we find its derivatives: \(y' = mx^{m-1}\) \(y'' = m(m-1)x^{m-2}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ x^2 [m(m-1)x^{m-2}] + x [mx^{m-1}] - [x^m] = 0 \] \[ m(m-1)x^{m-2}x^2 + mx^{m-1}x^1 - x^m = 0 \] \[ m(m-1)x^m + mx^m - x^m = 0 \] Factor out \(x^m\) (since \(x > 0\), \(x^m \neq 0\)): \[ x^m [m(m-1) + m - 1] = 0 \] Since \(x^m \neq 0\), the auxiliary (or characteristic) equation is: \[ m(m-1) + m - 1 = 0 \]

Step 3: Solve the Auxiliary Equation

Expand and simplify the auxiliary equation: \[ m^2 - m + m - 1 = 0 \] \[ m^2 - 1 = 0 \] This is a quadratic equation. We can solve it by factoring as a difference of squares: \[ (m-1)(m+1) = 0 \] This gives two distinct real roots: \[ m_1 = 1 \quad \text{and} \quad m_2 = -1 \]

Step 4: Write the General Solution based on the Roots

For a Cauchy-Euler equation with two distinct real roots \(m_1\) and \(m_2\), the general solution is of the form: \[ y = c_1 x^{m_1} + c_2 x^{m_2} \] Substitute \(m_1 = 1\) and \(m_2 = -1\): \[ y = c_1 x^1 + c_2 x^{-1} \] \[ y = c_1 x + \frac{c_2}{x} \] where \(c_1\) and \(c_2\) are arbitrary constants.

Thus, the general solution to the differential equation is \(y = c_1 x + \frac{c_2}{x}\).

This is a homogeneous linear second-order Cauchy-Euler differential equation. Note that the \(xy'\) term is missing, meaning its coefficient is 0.

Step 1: Assume a Solution Form

Assume a solution of the form \(y = x^m\). Then the derivatives are: \(y' = mx^{m-1}\) \(y'' = m(m-1)x^{m-2}\)

Step 2: Substitute into the Differential Equation

Substitute \(y\), \(y'\), and \(y''\) into the given differential equation: \[ 4x^2 [m(m-1)x^{m-2}] + 17 [x^m] = 0 \] \[ 4m(m-1)x^{m-2}x^2 + 17x^m = 0 \] \[ 4m(m-1)x^m + 17x^m = 0 \] Factor out \(x^m\) (assuming \(x \neq 0\)): \[ x^m [4m(m-1) + 17] = 0 \] Since \(x^m \neq 0\), the auxiliary (or characteristic) equation is: \[ 4m(m-1) + 17 = 0 \]

Step 3: Solve the Auxiliary Equation

Expand and simplify the auxiliary equation: \[ 4m^2 - 4m + 17 = 0 \] This is a quadratic equation. We use the quadratic formula \(m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\): Here, \(a=4\), \(b=-4\), \(c=17\). \[ m = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(4)(17)}}{2(4)} \] \[ m = \frac{4 \pm \sqrt{16 - 272}}{8} \] \[ m = \frac{4 \pm \sqrt{-256}}{8} \] \[ m = \frac{4 \pm \sqrt{256}i}{8} \] \[ m = \frac{4 \pm 16i}{8} \] \[ m = \frac{4}{8} \pm \frac{16}{8}i \] \[ m = \frac{1}{2} \pm 2i \] So, the roots are complex conjugates of the form \(m = \alpha \pm i\beta\), where \(\alpha = \frac{1}{2}\) and \(\beta = 2\).

Step 4: Write the General Solution based on Complex Roots

For a Cauchy-Euler equation with complex conjugate roots \(m = \alpha \pm i\beta\), the general solution is of the form: \[ y = x^{\alpha}[c_1 \cos(\beta \ln|x|) + c_2 \sin(\beta \ln|x|)] \] Substituting \(\alpha = \frac{1}{2}\) and \(\beta = 2\): \[ y(x) = x^{1/2}[c_1 \cos(2 \ln|x|) + c_2 \sin(2 \ln|x|)] \] Since the problem implicitly uses \(x=1\) in the initial conditions, we assume \(x > 0\) and can write \(\ln x\): \[ y(x) = \sqrt{x}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] \]

Step 5: Apply Initial Conditions to Find \(c_1\) and \(c_2\)

We have \(y(1) = -1\) and \(y'(1) = -\frac{1}{2}\).

First, use \(y(1) = -1\): \[ y(1) = \sqrt{1}[c_1 \cos(2 \ln 1) + c_2 \sin(2 \ln 1)] = -1 \] Since \(\ln 1 = 0\), \(\cos(0) = 1\), and \(\sin(0) = 0\): \[ 1[c_1 (1) + c_2 (0)] = -1 \] \[ c_1 = -1 \]

Next, we need \(y'(x)\). Differentiate \(y(x) = x^{1/2}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)]\) using the product rule: \(y' = \frac{d}{dx}(x^{1/2})[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] + x^{1/2}\frac{d}{dx}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)]\)

\(\frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2}\) \(\frac{d}{dx}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] = c_1(-\sin(2 \ln x) \cdot \frac{2}{x}) + c_2(\cos(2 \ln x) \cdot \frac{2}{x})\) \(= -\frac{2c_1}{x}\sin(2 \ln x) + \frac{2c_2}{x}\cos(2 \ln x)\)

So, \[ y'(x) = \frac{1}{2}x^{-1/2}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] + x^{1/2}\left(-\frac{2c_1}{x}\sin(2 \ln x) + \frac{2c_2}{x}\cos(2 \ln x)\right) \] Simplify the second term by combining \(x^{1/2}\) and \(\frac{1}{x} = x^{-1}\): \(x^{1/2}x^{-1} = x^{-1/2}\). \[ y'(x) = \frac{1}{2}x^{-1/2}[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] + x^{-1/2}[-2c_1 \sin(2 \ln x) + 2c_2 \cos(2 \ln x)] \] Factor out \(x^{-1/2}\): \[ y'(x) = x^{-1/2} \left[ \frac{1}{2}(c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)) -2c_1 \sin(2 \ln x) + 2c_2 \cos(2 \ln x) \right] \]

Now, use \(y'(1) = -\frac{1}{2}\). At \(x=1\), \(\ln 1 = 0\), \(\cos(0)=1\), \(\sin(0)=0\): \[ y'(1) = 1^{-1/2} \left[ \frac{1}{2}(c_1 (1) + c_2 (0)) -2c_1 (0) + 2c_2 (1) \right] = -\frac{1}{2} \] \[ 1 \left[ \frac{1}{2}c_1 + 2c_2 \right] = -\frac{1}{2} \] Substitute \(c_1 = -1\): \[ \frac{1}{2}(-1) + 2c_2 = -\frac{1}{2} \] \[ -\frac{1}{2} + 2c_2 = -\frac{1}{2} \] \[ 2c_2 = 0 \] \[ c_2 = 0 \]

Step 6: Write the Particular Solution

Substitute the values of \(c_1 = -1\) and \(c_2 = 0\) back into the general solution: \[ y(x) = \sqrt{x}[(-1) \cos(2 \ln x) + (0) \sin(2 \ln x)] \] \[ y(x) = -\sqrt{x} \cos(2 \ln x) \]

Thus, the particular solution to the differential equation with the given initial conditions is \(y(x) = -\sqrt{x} \cos(2 \ln x)\).

This is a non-homogeneous linear second-order Cauchy-Euler differential equation. We will solve it using the method of Variation of Parameters.

Step 1: Solve the Homogeneous Equation (\(y_h\))

First, consider the associated homogeneous equation: \[ x^2y'' - 3xy' + 3y = 0 \] Assume a solution of the form \(y = x^m\). Substitute \(y=x^m\), \(y'=mx^{m-1}\), and \(y''=m(m-1)x^{m-2}\) into the homogeneous equation: \[ x^2 [m(m-1)x^{m-2}] - 3x [mx^{m-1}] + 3 [x^m] = 0 \] \[ m(m-1)x^m - 3mx^m + 3x^m = 0 \] Factor out \(x^m\): \[ x^m [m(m-1) - 3m + 3] = 0 \] The auxiliary equation is: \[ m^2 - m - 3m + 3 = 0 \] \[ m^2 - 4m + 3 = 0 \] Factor the quadratic equation: \[ (m-1)(m-3) = 0 \] This gives two distinct real roots: \[ m_1 = 1 \quad \text{and} \quad m_2 = 3 \] The complementary solution (homogeneous solution) is: \[ y_h = c_1 x^{m_1} + c_2 x^{m_2} = c_1 x + c_2 x^3 \] From this, we identify two linearly independent solutions for the homogeneous equation: \(y_1(x) = x\) \(y_2(x) = x^3\)

Step 2: Convert to Standard Form

The method of Variation of Parameters requires the differential equation to be in the standard form \(y'' + P(x)y' + Q(x)y = f(x)\). Divide the original non-homogeneous equation by \(x^2\): \[ y'' - \frac{3}{x}y' + \frac{3}{x^2}y = 2x^2 e^x \] From this, we identify \(f(x) = 2x^2 e^x\).

Step 3: Calculate the Wronskian \(W(y_1, y_2)\)

The Wronskian of \(y_1\) and \(y_2\) is given by: \[ W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1 y_2' - y_2 y_1' \] We have \(y_1 = x\) and \(y_2 = x^3\). \(y_1' = 1\) \(y_2' = 3x^2\) \[ W(x, x^3) = x(3x^2) - x^3(1) = 3x^3 - x^3 = 2x^3 \]

Step 4: Calculate \(W_1\) and \(W_2\)

\(W_1\) is obtained by replacing the first column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_1 = \begin{vmatrix} 0 & y_2 \\ f(x) & y_2' \end{vmatrix} = 0 \cdot y_2' - y_2 \cdot f(x) = -y_2 f(x) \] \[ W_1 = -x^3 (2x^2 e^x) = -2x^5 e^x \]

\(W_2\) is obtained by replacing the second column of the Wronskian matrix with \(\begin{pmatrix} 0 \\ f(x) \end{pmatrix}\): \[ W_2 = \begin{vmatrix} y_1 & 0 \\ y_1' & f(x) \end{vmatrix} = y_1 \cdot f(x) - 0 \cdot y_1' = y_1 f(x) \] \[ W_2 = x (2x^2 e^x) = 2x^3 e^x \]

Step 5: Find \(u_1'\) and \(u_2'\)

The derivatives of the functions \(u_1(x)\) and \(u_2(x)\) are given by: \[ u_1' = \frac{W_1}{W} = \frac{-2x^5 e^x}{2x^3} = -x^2 e^x \] \[ u_2' = \frac{W_2}{W} = \frac{2x^3 e^x}{2x^3} = e^x \]

Step 6: Integrate \(u_1'\) and \(u_2'\) to Find \(u_1\) and \(u_2\)

Integrate \(u_1'\): \[ u_1 = \int -x^2 e^x \, dx \] This requires integration by parts. Let \(I = \int x^2 e^x \, dx\). \(\int u \, dv = uv - \int v \, du\) Let \(u_{int} = x^2\), \(dv_{int} = e^x \, dx\). Then \(du_{int} = 2x \, dx\), \(v_{int} = e^x\). \(I = x^2 e^x - \int 2x e^x \, dx\) Now integrate \(\int 2x e^x \, dx\). Let \(u_{int} = 2x\), \(dv_{int} = e^x \, dx\). Then \(du_{int} = 2 \, dx\), \(v_{int} = e^x\). \(\int 2x e^x \, dx = 2x e^x - \int 2 e^x \, dx = 2x e^x - 2e^x\) So, \(I = x^2 e^x - (2x e^x - 2e^x) = x^2 e^x - 2x e^x + 2e^x = e^x(x^2 - 2x + 2)\). Therefore: \[ u_1 = -e^x(x^2 - 2x + 2) \]

Integrate \(u_2'\): \[ u_2 = \int e^x \, dx = e^x \]

Step 7: Form the Particular Solution \(y_p\)

The particular solution is given by \(y_p = u_1 y_1 + u_2 y_2\): \[ y_p = (-e^x(x^2 - 2x + 2)) (x) + (e^x) (x^3) \] \[ y_p = -x e^x(x^2 - 2x + 2) + x^3 e^x \] \[ y_p = -x^3 e^x + 2x^2 e^x - 2x e^x + x^3 e^x \] \[ y_p = 2x^2 e^x - 2x e^x \]

Step 8: Write the General Solution

The general solution is the sum of the complementary solution and the particular solution: \[ y(x) = y_h(x) + y_p(x) \] \[ y(x) = c_1 x + c_2 x^3 + 2x^2 e^x - 2x e^x \]

Thus, the general solution to the differential equation is \(y = c_1 x + c_2 x^3 + 2x^2 e^x - 2x e^x\).

For Repeated Roots of Multiplicity K:

If an auxiliary equation has a root \(m\) with multiplicity \(K\), then the corresponding linearly independent solutions are: \(x^m, x^m \ln|x|, x^m (\ln|x|)^2, \ldots, x^m (\ln|x|)^{K-1}\).

Reduction to Constant Coefficients:

For a Cauchy-Euler equation, it can be transformed into a linear differential equation with constant coefficients by the substitution: Let \(x = e^t\) (so \(t = \ln|x|\)). Then solve the transformed equation for \(y(t)\), and finally substitute \(t = \ln|x|\) back to get \(y(x)\).

Different Form of Cauchy-Euler Equation:

A generalized form of the Cauchy-Euler equation is: \(a(x-x_0)^2 \frac{d^2 y}{dx^2} + b(x-x_0) \frac{dy}{dx} + c y = 0\)

Method of Solution:

  1. Substitution: Let \(t = x-x_0\). Then \(dt = dx\). The derivatives transform as: \(\frac{dy}{dx} = \frac{dy}{dt} \frac{dt}{dx} = \frac{dy}{dt}\) \(\frac{d^2 y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d}{dt}\left(\frac{dy}{dt}\right)\frac{dt}{dx} = \frac{d^2 y}{dt^2}\) The equation becomes: \(a t^2 \frac{d^2 y}{dt^2} + b t \frac{dy}{dt} + c y = 0\), which is a standard Cauchy-Euler equation in terms of \(t\).
  2. Directly solve: Assume a solution of the form \(y = (x-x_0)^m\). Then: \(y' = m(x-x_0)^{m-1}\) \(y'' = m(m-1)(x-x_0)^{m-2}\) Substitute these into the equation to obtain the auxiliary equation \(am(m-1) + bm + c = 0\), just like the standard form. The solutions will then be in terms of \((x-x_0)\).