Solve the differential equation by variation of parameters, subject to the initial conditions $y(0) = 1$, $y'(0) = 0$: $36y'' - y = x e^{\frac{x}{6}}$

Find $y(x) =$

Neetesh Kumar | October 27, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec06 (Homework) Question - 3
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Step-by-step solution:

To solve the differential equation $36y'' - y = x e^{\frac{x}{6}}$ with initial conditions $y(0) = 1$ and $y'(0) = 0$, we proceed with the following steps.

Step 1: Rewrite the Differential Equation

Rewrite the equation in standard form: $y'' - \frac{1}{36} y = \frac{x}{36} e^{\frac{x}{6}}$

Now, we have: $y'' - \frac{1}{36} y = \frac{x}{36} e^{\frac{x}{6}}$

Step 2: Solve the Homogeneous Equation

The associated homogeneous equation is: $y'' - \frac{1}{36} y = 0$

The characteristic equation for this differential equation is: $r^2 - \frac{1}{36} = 0$

Solving for $r$, we get: $r = \pm \frac{1}{6}$

Thus, the general solution to the homogeneous equation is: $y_h(x) = C_1 e^{\frac{x}{6}} + C_2 e^{-\frac{x}{6}}$

Step 3: Apply Variation of Parameters

We seek a particular solution of the form: $y_p(x) = u_1(x) e^{\frac{x}{6}} + u_2(x) e^{-\frac{x}{6}}$

where $u_1(x)$ and $u_2(x)$ are functions to be determined. For variation of parameters, we impose the conditions:

$u_1'(x) e^{\frac{x}{6}} + u_2'(x) e^{-\frac{x}{6}} = 0$

$\frac{1}{6} u_1'(x) e^{\frac{x}{6}} - \frac{1}{6} u_2'(x) e^{-\frac{x}{6}} = \frac{x}{36} e^{\frac{x}{6}}$

Step 4: Solve for $u_1'(x)$ and $u_2'(x)$

We have the system:

1. $u_1'(x) e^{\frac{x}{6}} + u_2'(x) e^{-\frac{x}{6}} = 0$
2. $\frac{1}{6} u_1'(x) e^{\frac{x}{6}} - \frac{1}{6} u_2'(x) e^{-\frac{x}{6}} = \frac{x}{36} e^{\frac{x}{6}}$

Multiply the second equation by 6 to simplify:

$u_1'(x) e^{\frac{x}{6}} - u_2'(x) e^{-\frac{x}{6}} = \frac{x}{6} e^{\frac{x}{6}}$

Now we can add and subtract these equations to solve for $u_1'(x)$ and $u_2'(x)$.

Adding the two equations: $2 u_1'(x) e^{\frac{x}{6}} = \frac{x}{6} e^{\frac{x}{6}}$

Thus, $u_1'(x) = \frac{x}{12}$

Subtracting the two equations: $2 u_2'(x) e^{-\frac{x}{6}} = -\frac{x}{6} e^{\frac{x}{6}}$

Thus, $u_2'(x) = -\frac{x}{12}e^{\frac{x}{3}}$

Step 5: Integrate to Find $u_1(x)$ and $u_2(x)$

Integrate $u_1'(x)$ and $u_2'(x)$ with respect to $x$:

1. $u_1(x) = \int \frac{x}{12} \, dx = \frac{x^2}{24}$
2. $u_2(x) = \int -\frac{x}{12}e^{\frac{x}{3}} \, dx = \frac{(3-x)e^{\frac{x}{3}}}{4}$

Step 6: Form the Particular Solution $y_p(x)$

Substitute $u_1(x)$ and $u_2(x)$ into $y_p(x) = u_1(x) e^{x/6} + u_2(x) e^{-x/6}$:

$y_p(x) = \frac{x^2}{24} e^{\frac{x}{6}} + \frac{(3-x)e^{\frac{x}{3}}}{4} e^{-\frac{x}{6}}$

Simplify the expression:

$y_p(x) = \frac{x^2}{24} e^{\frac{x}{6}} + \frac{(3-x)e^{\frac{x}{6}}}{4}$

Step 7: Write the General Solution

The general solution to the differential equation is:

$y(x) = y_h(x) + y_p(x)$

Substitute $y_h(x) = C_1 e^{\frac{x}{6}} + C_2 e^{-\frac{x}{6}}$ and $y_p(x) = \frac{x^2}{24} e^{\frac{x}{6}} + \frac{(3-x)e^{\frac{x}{6}}}{4}$:

$y(x) = C_1 e^{\frac{x}{6}} + C_2 e^{-\frac{x}{6}} + \frac{x^2}{24} e^{\frac{x}{6}} + \frac{(3-x)e^{\frac{x}{6}}}{4}$

After simplifying further

$y(x) = C_1 e^{\frac{x}{6}} + C_2 e^{-\frac{x}{6}} + (\frac{x^2-6x+18}{24} )e^{\frac{x}{6}}$

Step 8: Apply Initial Conditions

Given $y(0) = 1$ and $y'(0) = 0$, we will determine $C_1$ and $C_2$.

1. Apply $y(0) = 1$:

   $y(0) = C_1 e^0 + C_2 e^0 + \frac{3}{4} = 1$

   So, $C_1 + C_2 = \frac{1}{4}$.

2. Apply $y'(0) = 0$:

   First, find $y'(x)$:

   $y'(x) = \frac{1}{6} C_1 e^{\frac{x}{6}} - \frac{1}{6} C_2 e^{-\frac{x}{6}} + (\frac{x^2-6x+18}{144} )e^{\frac{x}{6}} + (\frac{2x-6}{24} )e^{\frac{x}{6}}$

   Substitute $x = 0$:

   $y'(0) = \frac{1}{6} C_1 - \frac{1}{6} C_2 - \frac{1}{8} = 0$

   This gives $4C_1 - 4C_2 = 3$.

   From $4C_1 + 4C_2 = 1$, we find $C_1 = \frac{1}{2}, C_2 = -\frac{1}{4}$.

Final Answer

The solution is:

$y(x) = \boxed{\frac{1}{2} e^{\frac{x}{6}} - \frac{1}{4} e^{-\frac{x}{6}} + \bigg(\frac{x^2-6x+18}{24}\bigg)e^{\frac{x}{6}}}$

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