Question

Solve each differential equation by variation of parameters

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{2}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

$y=c_1{e}^x+c_{2}x{e}^x-\frac{1}{2}\ln \left(1+x^2\right)e^x+\left({\tan }^{-1}x\right)x{e}^x$

Step-by-step solution

Step 1: Determine the corresponding homogeneous equation

First: we have to obtain the solution$\left(y_h\right)$ to the associated homogeneous equation:$y^{\text{'}\text{'}}-2y^{\text{'}}+y=0$

let $y=e^{\lambda x}$ ……. (1)

then: $y^{\text{'}}=\lambda e^{\lambda x}$ ……. (2)

$y^{\text{'}\text{'}}={\lambda }^2{e}^{\lambda x}$ ……. (3)

Substitute (1), (2), (3) into the homogeneous equation:

$\begin{array}{rcl}{\lambda }^2{e}^{\lambda x}-2\lambda e^{\lambda x}+e^{\lambda x}& =& 0\left(÷e^{\lambda x}\right)\\ {\lambda }^2-2\lambda +1& =& 0\\ (\lambda -1)(\lambda -1)& =& 0\\ \lambda & =& 1,1\end{array}$

$y_h=c_1{e}^x+c_{2}x{e}^x$

Determine the general solution for given differential equation

$y_1=e^x{y}_2=x{e}^x$

"i had to add an extra x to$y_2$ in order to avoid repetition"

$W_{\left(y_1,y_2\right)}=\left|\begin{array}{cc}{y}_1& y_2\\ \frac{d{y}_1}{dx}& \frac{d{y}_2}{dx}\end{array}\right|$

let$y_p=y_1{v}_1+y_2{v}_2$

$\begin{array}{rcl}{v}_1& =& \int \frac{y_{2}f\left(x\right)}{W\left(y_1,y_2\right)}\\ v_1& =& \int \frac{x{e}^x\frac{e^x}{1+x^2}}{e^{2x}}\\ v_1& =& \int \frac{x}{1+x^2}\end{array}$

$\begin{array}{rcl}{v}_1& =& -\frac{1}{2}\int \frac{-2x}{1+x^2}\\ v_1& =& -\frac{1}{2}\ln \left(1+x^2\right)\\ v_2& =& \int \frac{y_{1}f\left(x\right)}{W\left(y_1,y_2\right)}\\ v_2& =& \int \frac{e^x\frac{e^x}{1+x^2}}{e^{2x}}\end{array}$

$v_2=\int \frac{1}{1+x^2}m$(m identity integral)

$\begin{array}{rcl}{v}_2& =& {\tan }^{-1}x{y}_p\\ & =& -\frac{1}{2}\ln \left(1+x^2\right)e^x+\left({\tan }^{-1}x\right)x{e}^{x}y\\ & =& c_1{e}^x+c_{2}x{e}^x-\frac{1}{2}\ln \left(1+x^2\right)e^x+\left({\tan }^{-1}x\right)x{e}^x\end{array}$

$y=c_1{e}^x+c_{2}x{e}^x-\frac{1}{2}\ln \left(1+x^2\right)e^x+\left({\tan }^{-1}x\right)x{e}^x$