Solve $(D^3+3D^2+2D)y=x^2, D \equiv \dfrac{d}{dx}$
Choosing, trial solution as $y=a_0+a_1x+a_2x^2$
If i substitute this in given differential equation,
$0+3(2a_2)+2(a_1+2a_2x)=x^2 $
Comparing corresponding coefficients, i get,
$a_0=0=a_1 = a_2$ –> it seems i made wrong some where, Pls correct me.
Best Answer
Hint: Try with $y_p=A+B x+C x^2+D x^3$ then $$(D^3+3D^2+2D)y=(2 B + 6 C + 6 D)+ (4 C + 18 D)x+ 6 Dx^2\equiv x^2$$