Answer to Question #238375 in Differential Equations for Juwel

Question

X^2d^2y/dx^2-3xdy/dx+4y=x+x^2logx

Answer 1

Let us solve the differential equation $x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+4y=x+x^2\ln x.$ For this let us use the transformation $x=e^t.$ Then $y'_x=y'_te^{-t},\ y''_{x^2}=(y''_{t^2}-y'_t)e^{-2t}.$

Then we get the following equation

$e^{2t}(y''_{t^2}-y'_t)e^{-2t}-3e^ty'_te^{-t}+4y=e^t+te^{2t},$ which is equivalent to $y''_{t^2}-4y'_t+4y=e^t+te^{2t}.$

The characteristic equation $k^2-4k+4=0$ of the equation $y''_{t^2}-4y'_t+4y=0$ is equivalent to $(k-2)^2=0,$ and hence has the solutions $k_1=k_2=2.$

It follows that the general solution of the equation $y''_{t^2}-4y'_t+4y=e^t+te^{2t}$ is $y(t)=(C_1+C_2t)e^{2t}+y_p,$ where $y_p=ae^t+t^2(bt+c)e^{2t}.$

Then

$y'_p(t)=ae^t+(3bt^2+2ct)e^{2t}+(bt^3+ct^2)2e^{2t}= ae^t+(2bt^3+(3b+2c)t^2+2ct)e^{2t},$

$y''_p(t)=ae^t+(6bt^2+2(3b+2c)t+2c)e^{2t}+2(2bt^3+(3b+2c)t^2+2ct)e^{2t}=\\ ae^t+(4bt^3+(12b+4c)t^2+(6b+6c)t+2c)e^{2t}.$

Then we get $ae^t+(4bt^3+(12b+4c)t^2+(6b+6c)t+2c)e^{2t}-4(ae^t+(2bt^3+(3b+2c)t^2+2ct)e^{2t})+4(ae^t+(bt^3+ct^2)e^{2t})=e^t+te^{2t}.$

It follows that

$ae^t+((6b-2c)t+2c)e^{2t}=e^t+te^{2t}.$

Then we get $a=1,\ 6b-2c=1,\ 2c=0,$ and hence $a=1,\ b=\frac{1}6,\ c=0.$

Therefore, the solution of $y''_{t^2}-4y'_t+4y=e^t+te^{2t}$ is

$y(t)=(C_1+C_2t)e^{2t}+e^t+\frac{1}6t^3e^{2t}.$

Consequently, the solution of $x^2\frac{d^2y}{dx^2}-3x\frac{dy}{dx}+4y=x+x^2\ln x$ is

$y(x)=(C_1+C_2\ln x)x^2+x+\frac{1}6x^2\ln^3x.$