Solve the given differential equation by variation of parameters: $x^2 y'' - x y' + y = 8x$

Neetesh Kumar | October 28, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh4Sec07 (Homework) Question - 5
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Step-by-step solution:

This is a Cauchy-Euler differential equation, where the general form is:

$x^2 y'' + a x y' + b y = g(x)$

In this case, the equation is: $x^2 y'' - x y' + y = 8x$

Step 1: Solve the Homogeneous Equation

The corresponding homogeneous equation is: $x^2 y'' - x y' + y = 0$

For Cauchy-Euler equations, we assume a solution of the form $y = x^r$.

Substitute $y = x^r$ into the Homogeneous Equation

If $y = x^r$, then: $y' = r x^{r-1}$ $y'' = r(r - 1) x^{r - 2}$

Substitute these into the homogeneous equation: $x^2 \cdot r(r - 1) x^{r - 2} - x \cdot r x^{r - 1} + x^r = 0$

This simplifies to: $r(r - 1) x^r - r x^r + x^r = 0$

Factor out $x^r$: $x^r (r^2 - 2r + 1) = 0$

Since $x^r \neq 0$, the characteristic equation is: $r^2 - 2r + 1 = 0$

Solve the Characteristic Equation

The characteristic equation $r^2 - 2r + 1 = 0$ can be factored as: $(r - 1)^2 = 0$

This gives a repeated root: $r = 1$

General Solution to the Homogeneous Equation

With a repeated root $r = 1$, the solution to the homogeneous equation is: $y_h(x) = C_1 x + C_2 x \ln x$

Step 2: Apply Variation of Parameters to Find a Particular Solution

For the non-homogeneous equation $x^2 y'' - x y' + y = 8x$, we use 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 $y_1(x) = x$ and $y_2(x) = x \ln x$ are the solutions of the homogeneous equation.

Finding $u_1(x)$ and $u_2(x)$ by Cramer's Rule

We impose the conditions for variation of parameters:

1. $u_1'(x) y_1(x) + u_2'(x) y_2(x) = 0$
2. $u_1'(x) y_1'(x) + u_2'(x) y_2'(x) = \frac{8x}{x^2} = \frac{8}{x}$

Calculate the Wronskian

The solutions to the homogeneous equation are $y_1(x) = x$ and $y_2(x) = xln(x)$.

The Wronskian $W(y_1, y_2)$ is given by: $W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = \begin{vmatrix} x & xln(x) \\ 1 & 1+ln(x) \end{vmatrix}$

Calculating this determinant: $W(y_1, y_2) = x$

Find $u_1'(x)$ and $u_2'(x)$ using Cramer's Rule

Using Cramer’s rule, we find:

1. For $u_1'(x)$:

   $u_1'(x) = \frac{1}{W} \begin{vmatrix} 0 & xln(x) \\ \frac{8}{x} & 1+ln(x) \end{vmatrix}$

   Simplifying, we find $u_1'(x) = \frac{-8ln(x)}{x}$.

2. For $u_2'(x)$:

   $u_2'(x) = \frac{1}{W} \begin{vmatrix} x & 0 \\ 1 & \frac{8}{x} \end{vmatrix}$

   Simplifying, we find $u_2'(x) = \frac{8}{x}$.

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

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

1. $u_1(x) = \int \frac{-8ln(x)}{x} \, dx = -4(ln(x))^2$

   
   

2. $u_2(x) = \int \frac{8}{x} \, dx = 8ln(x)$

Step 5: Form the Particular Solution $y_p(x)$

Substitute $u_1(x)$ and $u_2(x)$ into $y_p(x) = u_1(x).x + u_2(x).xln(x)$:

$y_p(x) = -4x(ln(x))^2 + 8x(ln(x))^2 = 4(ln(x))^2$

Step 6: Write the General Solution

The general solution to the differential equation is:

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

Substitute $y_h(x) = C_1.x + C_2.xln(x)$ and $y_p(x) = -4(ln(x))^2$ :

$y(x) = C_1.x + C_2.xln(x) -4(ln(x))^2$

Final Answer

$y(x) = \boxed{C_1.x + C_2.xln(x) -4(ln(x))^2}$

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