Answer to Question #215485 in Differential Equations for Krk

Question

(D²-2D+3) y=x²

Answer 1

Let us solve the differential equation $(D^2-2D+3) y=x^2.$ Its characteristic equation $k^2-2k+3=0$ is equivalent to $(k-1)^2=-2,$ and hence has the roots $k_1=1+i\sqrt{2}$ and $k_2=1-i\sqrt{2}.$ Therefore, the general solution is of the form $y=e^x(C_1\cos(\sqrt{2}x)+C_2\sin(\sqrt{2}x))+y_p,$ where $y_p=ax^2+bx+c.$ It follows that $y_p'=2ax+b,\ y_p''=2a.$ We have the equation $2a-2(2ax+b)+3(ax^2+bx+c)=x^2,$ which is equivalent to $3ax^2+(-4a+3b)x+(-2b+3c)=x^2.$ It follows that $3a=1,\ -4a+3b=0,\ -2b+3c=0.$ Then $a=\frac{1}{3},\ b=\frac{4}{9},\ c=\frac{8}{27}.$ We conclude that the general solution is the following:

$y=e^x(C_1\cos(\sqrt{2}x)+C_2\sin(\sqrt{2}x))+\frac{1}{3}x^2+\frac{4}{9}x+\frac{8}{27}.$