$y'' + 3y' = 2x^4 + x^2e^{ -3x}+\sin3x$, Using Method of Undetermined Coefficients (Annihilator approach). I Found these two questions in one of my old past papers. I tried to attempt in an exam but failed.
I still haven't found the solution for these so if anyone is willing to solve them, i will give them my regards.
Will Someone help me get my revenge on these 2 questions? Vengeance is Sweet :D
Q) For the following set of four ODE’s determine suitable form of the particular solution ONLY. Do not evaluate co-efficient.

*

*$y''+3y'=2x^4 + x^2e^{-3x}+\sin3x$, using method of undetermined
coefficients (annihilator approach).


*$y'+2y'=3e^{-x}+2\cos xe^{-x}+4x^2e^{-x}+\sin x$, using method of undetermined coefficients (annihilator approach).
 A: $y''+3y'=0$ has roots $0,-3$ so $y=Ae^{0x}+Be^{-3x}=A+Be^{-3x}$
Since in RHS we have:

*

*$x^4=x^4e^{0x}$ which collides with root $0$ search for a particular solution $p(x)$ with $p$ polynomial of degree $5=4+1$.


*$x^2e^{-3x}$ collides with root $-3$ search for a particular solution $q(x)e^{-3x}$ with $q$ polynomial of degree $3=2+1$


*$\sin(3x)=\Im(e^{3ix})$ and $3i$ do not collide, so just search for a solution of same degree (i.e constant), that is $a\sin(3x)+b\cos(3x)$
A: First solve $y''+3y'=0$ taking $y=e^{mx} \implies m=0, m=-3 \implies y_1=1, y_2= e^{-3x}$ The wronskian of $y_1,y_2$ is $W(x)=-3e^{-3x}$ By variation of parameters the Solution of
$Y''+3Y'=f(x)$ is $$Y=C_1(x) + C_2(x) e^{-3x}, C_1(x)=D_1-\int \frac{e^{-3x} f(x)}{W(x)}dx, C_2=D_2+ \int \frac{1. f(x)}{W(x)} dx$$ By insertinf $f(x)=2x^4+x^2 e^{-3x}+\sin 3x$
and $W(x)$ we get
$$C_1(x)=D_1+\frac{1}{3}\int (2x^4+x^2 e^{-3x}+\sin 3x) dx,$$ $$C_2(x)=D_2-\frac{1}{3} \int (2x^4+x^2 e^{-3x}+\sin 3x) e^{3x} dx. $$
Carrying out these integrals and substituting yhem in $Y(x)$ we get the total solution of the ODE.
