Answer to Question #288877 in Differential Equations for Pankaj

Question

Solve :

x²y" -2xy' -4y=x²+2ln x

Answer 1

Let us solve the differential equation $x^2y'' -2xy' -4y=x^2+2\ln x.$

Let us use the transformation $x=e^t.$ Then $y'_x=y'_te^{-t},\ y''_{x^2}=(y''_{t^2}-y'_t)e^{-2t}.$

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

The characteristic equation $k^2-3k-4=0$ of the differential equation $y''_{t^2}-3y'_t -4y=0$ is equivalent to $(k+1)(k-4)=0,$ and hence has the roots $k_1=-1,\ k_2=4.$

It follows that the general solution of the differential equation $y''_{t^2}-3y'_t -4y=e^{2t}+2t$ is $y(t)=C_1e^{-t}+C_2e^{4t}+y_p,$ where $y_p=ae^{2t}+bt+d.$ Then $y_p'=2ae^{2t}+b,\ y_p''=4ae^{2t}.$

It follows that

$4ae^{2t}-3(2ae^{2t}+b)-4(ae^{2t}+bt+d)=e^{2t}+2t,$ and hence

$-6ae^{2t}-4bt-3b-4d=e^{2t}+2t.$

We conclude that $-6a=1,\ -4b=2,\ -3b-4d=0,$ and thus $a=-\frac{1}6,\ b=-\frac{1}2,\ d=-\frac{3}4b=\frac{3}8.$

Therefore, the general solution of the differential equation $y''_{t^2}-3y'_t -4y=e^{2t}+2t$ is $y(t)=C_1e^{-t}+C_2e^{4t}-\frac{1}6e^{2t}-\frac{1}2t+\frac{3}8.$

We conclude that the general solution of the differential equation $x^2y'' -2xy' -4y=x^2+2\ln x$ is $y(x)=C_1x^{-1}+C_2x^{4}-\frac{1}6x^{2}-\frac{1}2\ln x+\frac{3}8.$