Question

[C] Solve the following differential equation:

x^ 2 y^ " - 2xy'-4y=x^ 2 +2 log x

Answer 1

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

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

We get the equation $e^{2t}(y_{t^2}''-y_t')e^{-2t}- 2e^{t}y_t'e^{-t}-4y=e^{2t} +2t,$ 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$(k-4)(k+1)=0,$ and hence has the roots $k_1=4,\ k_2=-1.$

Therefore, the general solution of the last differential equation is $y(t)=C_1e^{4t}+C_2e^{-t}+y_p,$ where $y_p=ae^{2t}+bt+c.$ It follows that $y_p'=2ae^{2t}+b,\ y_p''=4ae^{2t},$ and hence we get

$4ae^{2t}-3(2ae^{2t}+b)-4(ae^{2t}+bt+c)= e^{2t} +2t.$

It follows that $-6ae^{2t}-4bt-3b-4c= e^{2t} +2t,$ and hence $-6a=-1,\ -4b=2,\ -3b-4c=0.$ Consequently, $a=\frac{1}6,\ b=-\frac{1}2,\ c=-\frac{3}4b=\frac{3}{8}.$

We conclude that $y(t)=C_1e^{4t}+C_2e^{-t}+\frac{1}6e^{2t}-\frac{1}2t+\frac{3}8.$

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