Question

Solve: $(D^2-4D+1)y=x^2$

Answer 1

We have to solve $(D^2-4D+1)y=x^2$

The characteristic equation is $p^2-4p+1=0$

$\Rightarrow p=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm 2\sqrt{3}}{2}=2\pm \sqrt{3}$

Thus Complementary function $C.F.=A{e}^{(2+\sqrt{3})x}+B{e}^{(2-\sqrt{3})x}$

Particular integral $P.I.=\frac{1}{D^2-4D+1}\left(x^2\right)$

$=[1-(4D-D^2\left){]}^{-1}\right(x^2)$

$=[1+(4D-D^2)+(4D-D^2{)}^2+...\left]\right(x^2)$

$=[1+4D+15D^2+...]\left(x^2\right)$

$\therefore P.I.=x^2+8x+30$

Hence the general solution is $y=C.F.+P.I.$

$y.=A{e}^{(2+\sqrt{3})x}+B{e}^{(2-\sqrt{3})x}+(x^2+8x+30)$