Question

Find the general solution of the ordinary differential equation (D2+2D+1)y=e^-xlnx using method of variation of parameters.

Answer 1

Let us find the general solution of the ordinary differential equation $(D^2+2D+1)y=e^{-x}\ln x$ using method of variation of parameters. The characteristic equation $k^2+2k+1=0$ of the homogeneous differential equation is equivalent to $(k+1)^2=0,$ and hence has the roots $k_1=k_2=-1.$ The general solution of the non-homogeneous equation is of the form $y=C_1(x)e^{-x}+C_2(x)xe^{-x}.$

Therefore, we should solve the following system

$\begin{cases}C_1'e^{-x}+C_2'xe^{-x}=0\\-C_1'e^{-x}+C_2'e^{-x}-C_2'xe^{-x}=e^{-x}\ln x \end{cases}$

which is equivalent to

$\begin{cases}C_1'=-C_2'x\\-C_1'+C_2'-C_2'x=\ln x \end{cases}$

$\begin{cases}C_1'=-C_2'x\\C_2'x+C_2'-C_2'x=\ln x \end{cases}$

$\begin{cases}C_1'=-C_2'x\\C_2'=\ln x \end{cases}$

$\begin{cases}C_1'=-x\ln x\\C_2'=\ln x \end{cases}$

It follows that

$C_1(x)=-\int x\ln xdx=$

|$u=\ln x, dv =xdx, du=\frac{dx}{x},v=\frac{x^2}{2}$ |

$=-\frac{1}{2}x^2\ln x+\frac{1}{2}\int xdx$

$=-\frac{1}{2}x^2\ln x+\frac{x^2}{4}+C_1$

$C_2(x)=\int\ln xdx=$

|$u=\ln x, dv =dx,du=\frac{dx}{x}, v=x$ |

$=x\ln x-\int dx$

$=x\ln x-x+C_2.$

We conclude that the general solution is

$y=(-\frac{1}{2}x^2\ln x+\frac{x^2}{4}+C_1)e^{-x}+(x\ln x-x+C_2)xe^{-x}$ or

$y=(\frac{1}{2}x^2\ln x-\frac{3}{4}x^2+C_1+C_2x)e^{-x}.$