Consider the following differential equation. $y'' + 16y' + 64y = e^{-8x}$ Proceed as in the example to find a particular solution $y_p(x)$ of the given differential equation in the integral form $y_p(x) = \int_{x_0}^x G(x, t) f(t) \, dt$.

Neetesh Kumar | October 29, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec08 (Homework) Question - 2
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To solve the differential equation $y'' + 16y' + 64y = e^{-8x}$ using variation of parameters (Wronskian method), we will first solve the associated homogeneous equation and then use the Wronskian to find the particular solution.

Step 1: Solve the Homogeneous Equation

The associated homogeneous equation is: $y'' + 16y' + 64y = 0$

The characteristic equation for this equation is: $r^2 + 16r + 64 = 0$

Solving for $r$: $r = \frac{-16 \pm \sqrt{256 - 256}}{2} = -8$

This gives a repeated root $r = -8$. Therefore, the solution to the homogeneous equation is: $y_h(x) = (C_1 + C_2 x) e^{-8x}$

From this, we can select two linearly independent solutions: $y_1(x) = e^{-8x}$ and $y_2(x) = x e^{-8x}$

Step 2: Compute the Wronskian of $y_1$ and $y_2$

The Wronskian $W(y_1, y_2)$ of two functions $y_1(x)$ and $y_2(x)$ is defined as:

$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}$

Compute $y_1'$ and $y_2'$:

$y_1' = -8 e^{-8x}$

$y_2' = e^{-8x} - 8x e^{-8x}$

Now, calculate the Wronskian:

$W(y_1, y_2) = \begin{vmatrix} e^{-8x} & x e^{-8x} \\ -8 e^{-8x} & (1 - 8x) e^{-8x} \end{vmatrix}$

Expanding this determinant:

$W(y_1, y_2) = e^{-8x} \cdot (1 - 8x) e^{-8x} - x e^{-8x} \cdot (-8 e^{-8x})$

$= e^{-16x} (1 - 8x) + 8x e^{-16x}$

$= e^{-16x}$

Thus, the Wronskian $W(y_1, y_2) = e^{-16x}$.

Step 3: Set Up Integrals for $u_1(x)$ and $u_2(x)$

Using variation of parameters, we assume a particular solution of the form:

$y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x)$

where $u_1(x)$ and $u_2(x)$ are functions to be determined.

According to variation of parameters, we have:
$u_1'(x) = -\frac{y_2(x) f(x)}{W(y_1, y_2)} = -\frac{x e^{-8x} \cdot e^{-8x}}{e^{-16x}} = -x$

$u_2'(x) = \frac{y_1(x) f(x)}{W(y_1, y_2)} = \frac{e^{-8x} \cdot e^{-8x}}{e^{-16x}} = 1$

Step 4: Integrate to Find $u_1(x)$ and $u_2(x)$

Now integrate $u_1'(x)$ and $u_2'(x)$ to find $u_1(x)$ and $u_2(x)$:

$u_1(x) = \int -x \, dx = -\frac{x^2}{2}$

$u_2(x) = \int 1 \, dx = x$

Step 5: Write the Particular Solution

The particular solution $y_p(x)$ from the Green's Function is then:

$y_p(x) = u_1(x) y_1(x) + u_2(x) y_2(x)$

$y_p(x) = \int_{x_0}^x \dfrac{(e^{-8t}.xe^{-8x}) - (e^{-8t}.te^{-8x})}{2} = \dfrac{(x-t)e^{-8(x+t)}}{2}.f(t)dt$

Step 6: Write the General Solution

The general solution to the differential equation is the sum of the homogeneous solution and the particular solution:

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

where

$y_h(x) = (C_1 + C_2 x) e^{-8x}$

and

$y_p(x) = \boxed{\dfrac{(x-t)e^{-8(x+t)}}{2}}.f(t)dt$

Final Answer

$y(x) = \boxed{C_1e^{-8x} + C_2 xe^{-8x}} + y_p(x)$

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