Having problem with the particular equation of 2nd order non-homogeneous differential equations

Question

Given these questions to solve on 2nd order non homogeneous equation. I'm having problem of forming the particular solution of it:
$$ \frac{d^2x}{dt^2} \,+4x=289te^t\,sin2t $$
$$ 2x\ddot y\,+\,\dot y \,-\,2y=0 $$
For the first equation, I got the homogeneous equation to be: $ y_c=C_1e^{2it} \,+C_2e^{-2it}$
after forming the particular equation $$ (A+Bt) e^t \sin2t + (C+Dt) e^t\cos2t $$
I applied method of undetermined coefficient but I got incorrect answer
Any idea of forming a better solution even without using method of undetermined coefficient

Answer 1

First equation, undetermined coefficients requires to see that the right side is not in resonance with the left side, thus the standard construct applies $$ x_p(t)=(A+Bt)e^t\sin(2t)+(C+Dt)e^t\cos(2t). $$

For the second equation the power series approach is called Frobenius method. The Euler-Cauchy part of the equation is $2x^2\ddot y+x\dot y=0$, so that the indicial equation $0=2m(m-1)+m=m(2m-1)$ gives basis solutions $1$ and $\sqrt x$, the power series are thus $$ y_1(x)=\sum_ka_kx^k~~\text{ and }~~y_2(t)=\sqrt x\sum_k b_kx^k. $$ Inserting and comparing coefficients should give the coefficient recursion for both.

---

Setting $y(x)=\sum a_kx^{k+r}$ with $y'(x)=\sum (k+r)a_kx^{k+r-1}$ and $y''(x)=\sum (k+r)(k+r-1)a_kx^{k+r-2}$ gives after comparing coefficients of equal degree terms $$ 2(k+r)(k+r-1)a_k+(k+r)a_k-2a_{k-1}=0\\~\\ (k+r)(k+r-\tfrac12)a_k=a_{k-1} $$ With $a_{-1}=0$ a non-trivial $a_0$ is only obtained for $r=0$ and $r=\frac12$. Then iterate \begin{align} r&=0:& a_k&=\frac{a_{k-1}}{k(k-\frac12)}&\implies~~&a_k=\frac{2^k}{(2k)!}a_0\\ r&=\tfrac12:&a_k&=\frac{a_{k-1}}{k(k+\frac12)}&\implies~~&a_k=\frac{2^k}{(2k+1)!}a_0 \end{align} Now compare with the power series of the hyperbolic functions.