Question

using the method of variation of parameters to determine the solution to the equation

d²y/dx²-6dy/dx+9y=e^3x/x³

Answer 1

Given the equation:

$$\frac{d^{2} y(x)}{d x^{2}}-6 \frac{d y(x)}{d x}+9 y(x)=\frac{e^{3 x}}{x^{3}} :$$

The general solution will be the sum of the complementary solution and particular solution. Find the complementary solution by solving

$$\frac{d^{2} y(x)}{d x^{2}}-6 \frac{d y(x)}{d x}+9 y(x)=0$$

Assume a solution will be proportional to $e^{\lambda x}$ for some constant $\lambda$ . Substitute $y(x)=e^{\lambda x}$ into the differential equation:

$$\frac{d^{2}}{d x^{2}}\left(e^{x x}\right)-6 \frac{d}{d x}\left(e^{\lambda x}\right)+9 e^{\lambda x}=0$$

Substitute

$$\frac{d^{2}}{d x^{2}}\left(e^{\lambda x}\right)=\lambda^{2} e^{\lambda x}$$

and

$$\frac{d}{d x}\left(e^{\lambda x}\right)=\lambda e^{\lambda x}$$

We have:

$$\lambda^{2} e^{\lambda x}-6 \lambda e^{\lambda x}+9 e^{\lambda x}=0$$

Factor out $\boldsymbol{e}^{\boldsymbol{\lambda x}}$ :

$$\left(\lambda^{2}-6 \lambda+9\right) e^{\lambda x}=0$$

Since $e^{\lambda x} \neq 0$ for any finite $\lambda$ , the zeros must come from the polynomial:

$$\lambda^2 - 6\lambda+9=0$$

Factor:

$$(\lambda-3)^2=0$$

Solving for $\lambda$ we get:

$$\lambda= 3 \text{ twice}$$

The general solution of the equation is:

$$y(x) = y_1(x) + y_2(x) = c_1e^{3x}+ c_2e^{3x}x$$

We determine the particular solution to

$$\frac{d^{2} y(x)}{d x^{2}}-6 \frac{d y(x)}{d x}+9 y(x)=\frac{e^{3 x}}{x^{3}}$$

by variation of parameters.

We list the basis solutions in $y_c(x)$ :

$$y_{b_{1}}(x) = e^{3x} \text{ and } y_{b_{2}}(x) = e^{3x} x$$

Compute the Wronskian of $y_{b_{1}}(x) \text{ and } y_{b_{2}}(x) :$

$$W(x)=\left|\begin{array}{cc}e^{3 x} & e^{3 x} x \\ \frac{d}{d x}\left(e^{3 x}\right) & \frac{d}{d x}\left(e^{3 x} x\right)\end{array}\right|=\left|\begin{array}{cc}e^{3 x} & e^{3 x} x \\ 3 e^{3 x} & e^{3 x}+3 e^{3 x} x\end{array}\right|=e^{6 x}$$

$\text{Let }f(x)=\frac{e^{3 x}}{x^{3}}:$

Let

$$v_{1}(x)=-\int \frac{f(x) y_{b_{2}}(x)}{w(x)} d x$$

and

$$v_{2}(x)=\int \frac{f(x) y_{b}(x)}{w(x)} d x$$

The particular solution will be given by:

$$y_{p}(x)=v_{1}(x) y_{b_{1}}(x)+v_{2}(x) y_{b_{2}}(x)$$

Compute $v_{1}(x)$ :

$$v_{1}(x)=-\int \frac{1}{x^{2}} d x=\frac{1}{x}$$

Compute $v_{2}(x)$ :

$$v_{2}(x)=\int \frac{1}{x^{3}} d x=-\frac{1}{2 x^{2}}$$

The particular solution is thus:

$$y_{p}(x)=v_{1}(x) y_{b_{1}}(x)+v_{2}(x) y_{b_{2}}(x)=\frac{e^{3 x}}{x}-\frac{e^{3 x}}{2 x}=\frac{e^{3 x}}{2 x}$$

The general solution is therefore:

$$y(x) = y_c(x) + y_p(x) = c_1e^{3x}+ c_2e^{3x}x +\frac{e^{3 x}}{2 x}$$