Question

(D^2-2D+3)y=X^2-1

Answer 1

Let us solve the differential equation $(D^2-2D+3)y=x^2-1$. For this let us solve the characteristic equation of the homogeneous equation $(D^2-2D+3)y=0$:

$k^2-2k+3=0$

$(k-1)^2+2=0$

$(k-1)^2=-2$

$k_1=1+i\sqrt{2}$ and $k_2=1-i\sqrt{2}.$

It follows that the general solution of the equation $(D^2-2D+3)y=x^2-1$ 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 $y_p'=2Ax+B$ and $y_p''=2A.$ Therefore, $2A-2(2Ax+B)+3(Ax^2+Bx+C)=x^2-1.$ We conclude that $3Ax^2+(3B-4A)x+(2A-2B+3C)=x^2-1.$ Consequently, $3A=1,\ 3B-4A=0,\ 2A-2B+3C=-1.$ It follows that $A=\frac{1}{3},\ B=\frac{4}{9},\ C=\frac{1}{3}(-1-\frac{2}{3}+\frac{8}{9})=-\frac{7}{27}.$

We conclude that the general solution of the differential equation $(D^2-2D+3)y=x^2-1$ is

$y=e^x(C_1\cos(\sqrt{2}x)+ C_2\sin(\sqrt{2}x))+\frac{1}{3}x^2+\frac{4}{9}x-\frac{7}{27}.$