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}

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Types of Solutions

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

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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.

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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.

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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

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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.

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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

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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.

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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.

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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).

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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.

2y'' - 5y' - 3y = 0

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}.

y'' - 10y' + 25y = 0

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}.

y'' + 4y' + 7y = 0

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}.

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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.

y'' + 3y' + 2y = 4x^2

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.

y'' - 3y' = 8e^{3x} + 4\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 (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.

y'' + y = x \cos x - \cos 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'' + 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.

y'' - 2y' + y = 10e^{-2x}\cos 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

y'' - 4y' + 4y = (x+1)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: 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}.

4y'' + 36y = \csc(3x)

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.

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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|)].

x^2y'' - 3xy' + 4y = 0, \quad x > 0

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.

x^2y'' + xy' - y = 0, \quad x > 0

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}.

4x^2y'' + 17y = 0, \quad y(1) = -1, \quad y'(1) = -\frac{1}{2}

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).

x^2y'' - 3xy' + 3y = 2x^4 e^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).