Question

Solve each differential equation by variation of parameters

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{9}\mathbf{=}\frac{\mathbf{9}\mathbf{x}}{{\mathbf{e}}^{\mathbf{3}\mathbf{x}}}\mathbf{=}\mathbf{9}\mathit{x}{\mathit{e}}^{\mathbf{-}\mathbf{3}\mathbf{x}}$

Short answer

$y\left(x\right)=c_1{e}^{3x}+c_2{e}^{-3x}-\frac{3x^2+x}{4}{e}^{-3x}$

Step-by-step solution

Step 1: Determine the particular solution for ycand yp  differential equation

The first step in obtaining the solution to

$y^{\text{'}\text{'}}-9=\frac{9x}{e^{3x}}=9x{e}^{-3x}$

is to form the complementary solution. To do that we first solve the complementary equation

$$\begin{gathered} m^2-9=0 \\ \Rightarrow (m-3)(m+3)=0 \end{gathered}$$

We see that the roots are$m_1=3$ and $m_2=-3$so$y_1\left(x\right)=e^{3x},y_2\left(x\right)=e^{-3x}$ and the complementary solution is given by

$y_c\left(x\right)=c_1{e}^{3x}+c_2{e}^{-3x}.$

We look for a particular solution with the form

$y_p\left(x\right)=u_1\left(x\right)y_1\left(x\right)+u_2\left(x\right)y_2\left(x\right)$

In order to find $u_1$and $u_2$we need the Wronskian $W_{,}{W}_{1}and{W}_2$:

$\begin{array}{rcl}W& =& \left|\begin{array}{cc}{e}^{3x}& e^{-3x}\\ 3e^{3x}& -3e^{-3x}\end{array}\right|=-3e^{3x}{e}^{-3x}-3e^{-3x}{e}^{3x}=-3-3=-6\\ W_1& =& \left|\begin{array}{cc}0& e^{-3x}\\ 9x{e}^{-3x}& -3e^{-3x}\end{array}\right|=-9x{e}^{-3x}{e}^{-3x}=-9x{e}^{-6x}\\ W& =& \left|\begin{array}{cc}{e}^{3x}& 0\\ 3e^{3x}& 9x{e}^{-3x}\end{array}\right|=9x{e}^{3x}{e}^{-3x}=9x\end{array}$

We can now determine $u{\text{'}}_{1}andu{\text{'}}_2$:

$$\begin{gathered} u_1^{\text{'}}=\frac{W_1}{W}=\frac{-9x{e}^{-6x}}{-6}=\frac{3}{2}x{e}^{6x}, \\ {u}_2^{\text{'}}=\frac{W_2}{W}=\frac{9x}{-6}=-\frac{3x}{2}x \end{gathered}$$

We get$u_1$and$u_2$by integrating (for$u_1$we need partial integration):

$$\begin{gathered} =-\frac{3}{2}\frac{x}{6}{e}^{-6x}+\frac{3}{2}\frac{1}{6}\int e^{-6x}dx \\ =-\frac{x}{4}{e}^{-6x}+\frac{1}{4}\int e^{-6x}dx \\ =-\frac{x{e}^{-6x}}{4}-\frac{1}{24}{e}^{-6x} \end{gathered}$$

$u_2\left(x\right)=-\frac{3}{2}\int xdx=-\frac{3}{2}\frac{x^2}{x}=-\frac{3x^2}{4}$

We conclude that

$\begin{array}{rcl}{y}_p\left(x\right)& =& -\left(\frac{x}{4}+\frac{1}{24}\right)e^{-6x}{e}^{3x}-\frac{3}{4}{x}^2{e}^{-3x}\\ & =& -\left(\frac{x}{4}+\frac{1}{24}\right)e^{-3x}-\frac{3}{4}{x}^2{e}^{-3x}\\ & =& -\frac{3x^2+x}{4}{e}^{-3x}-\frac{1}{24}{e}^{-3x}\end{array}$

Determine the general solution for given differential equation

and the general solution is equal to

$\begin{array}{rcl}y\left(x\right)& =& y_c\left(x\right)+y_p\left(x\right)\\ & =& c_1{e}^{3x}+c_2{e}^{-3x}-\frac{3x^2+x}{4}{e}^{-3x}-\frac{1}{24}{e}^{-3x}\end{array}$

Since$c_1$and$c_2$are arbitrary constants we can write y(x) as

$y\left(x\right)=c_1{e}^{3x}+c_2{e}^{-3x}-\frac{3x^2+x}{4}{e}^{-3x}$