Question

solve the given differential equation by variation of parameters.$\mathbf{}\mathbf{21}\mathbf{.}\mathbf{}{\mathit{x}}^{\mathbf{2}}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}$

Short answer

The solution for the given differential equation for variation of parameters

$y=c_{1}x+c_{2}x\ln x+x(\ln x{)}^2$

Step-by-step solution

Define Cauchy – Euler:

The Cauchy – Euler equation or Euler equation is simply a standard equation with a variable coefficient. It is sometimes called the equidimensional equation.

Find a homogenous differential equation:

$x^{2}y\text{'}\text{'}-xy\text{'}+y=2x$

Multiply both sides with $\frac{1}{x^2}$,

We get,

$y\text{'}\text{'}-\frac{1}{x}y\text{'}+\frac{1}{x^2}y=\frac{2}{x}$

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right),$

Consider,

$x^{2}y\text{'}\text{'}-xy\text{'}+y=0$→(1)

Let's consider,

$y=x^m$

$y\text{'}=m{x}^{(m-1)}$

$y\text{'}\text{'}=m(m-1)x^{(m-2)}$
Substitute in the above equation,

$\begin{array}{rcl}{x}^2\left[m(m-1)x^{(m-2)}\right]-x\left[m{x}^{(m-1)}\right]+x^m& =& 0\\ m(m-1)x^2{x}^{(m-2)}-m{x}^1{x}^{(m-1)}+x^m& =& 0\\ m(m-1)x^m-m{x}^m+x^m& =& 0\\ \left[m\right(m-1)-m+1]x^m& =& 0\\ \left(m^2-m-m+1\right)x^m& =& 0\\ \left(m^2-2m+1\right)x^m& =& 0\\ & & \end{array}$

Where x≠0,

$m^2-2m+1=0$

m$=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\begin{array}{rcl}m& =& \frac{-(-2)\pm \sqrt{{(-2)}^2-4\left(1\right)\left(1\right)}}{2\left(1\right)}\\ & =& \frac{2\pm \sqrt{4-4}}{2}\\ & =& \frac{2\pm \sqrt{0}}{2}\\ & =& \frac{2}{2}\\ & =& 1\\ & & \end{array}$

Hence, the homogenous differential equation

$y_h=c_{1}x+c_{2}x\ln x$→(2)

Use the Wronskian method:

Where $y_1=x$and $y_2=x\ln x$,

Using Differentials,

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y_1^{\text{'}}& y_2^{\text{'}}\end{array}\right|\\ & =& \left|\begin{array}{cc}x& x\ln x\\ 1& \ln x+1\end{array}\right|\\ & =& \left(x\right)\times (\ln x+1)-\left(x\ln x\right)\times \left(1\right)\\ & =& x\ln x+x-x\ln x\\ & =& x\end{array}$

As the right side of the differential equation $f\left(x\right)=\frac{2}{x}$

Then $W_1$and $W_2$,

From above we get $u_1^{\text{'}}$and $u_2^{\text{'}}$,

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W\left(y_1,y_2\right)}\\ & =& \frac{-2\ln x}{x}\end{array}$

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{2}{x}\end{array}$

Use integration for $u_1^{\text{'}}$ and $u_2^{\text{'}}$,

$v=\ln x$and $dv=\frac{1}{x}dx$

$\begin{array}{rcl}{u}_1& =& -2\int vdv\\ & =& -2\times \frac{v^2}{2}\\ & =& -v^2\\ & =& -(\ln x{)}^2\end{array}$

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dx\\ & =& \int \frac{2}{x}dx\\ & =& 2\ln x\\ & & \end{array}$

Find y:

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(-{\left(\ln x\right)}^2\right)\times \left(x\right)+\left(2\ln x\right)\times \left(x\ln x\right)\\ & =& -x(\ln x{)}^2+2x(\ln x{)}^2\\ & =& x(\ln x{)}^2 \dots \left(3\right)\\ & & \end{array}$

From the equation (2) and (3),

$y=y_h+y_p$

The solution is$y=c_{1}x+c_{2}x\ln x+x(\ln x{)}^2$