$\left\{X(t),t>0\right\}$ be a random variable and have $Y\sim U(0,\pi )$ distribution. Let $X(t)={e}^{2Y}t$. Find variance of $X(t)$
$Y\sim U(0,\pi )\Rightarrow {f}_{Y}(y)=\frac{1}{\pi }$ where $0\le y\le \pi$ then
$$ E\left[X(t)\right]=E\left[{e}^{2Y}t\right]=\int _{0}^{\pi }\frac{1}{\pi }{e}^{2y}tdy=t\frac{\left({e}^{2\pi }-1\right)}{2\pi } $$
$var\left[X(t)\right]=E\left[{X}^{2}(t)\right]-(E\left[X(t)\right]{)}^{2}=E\left[{e}^{4Y}{t}^{2}\right]-(\frac{{t}^{2}}{4{\pi }^{2}}({e}^{4\pi }-2{e}^{2\pi }+1))=\int _{0}^{\pi }\frac{1}{\pi }{e}^{4y }{t}^{2}dy-(\frac{{t}^{2}}{4{\pi }^{2}}({e}^{4\pi }-2{e}^{2\pi }+1))=\frac{{t}^{2}}{4\pi }\left({e}^{4\pi }-1\right)-(\frac{{t}^{2}}{4{\pi }^{2}}({e}^{4\pi }-2{e}^{2\pi }+1))$
Is my solution right? I feel like I'm missing something. Any idea will be appreciated.
Your solution is correct. You have done all that you need.
You make use of the fact that: $$\begin{align}\mathsf E(X^k(t))&=\mathsf E(t^k\mathrm e^{2kY})\\&=(t^k/\pi)\int_0^\pi \mathrm e^{2ky}\,\mathrm d y \\[1ex]&= (\mathrm e^{2k\pi}-1)(t^k/2k\pi)\\[2ex]\hspace{-50ex}\text{So}\\\mathsf E(X(t)) &= (\mathrm e^{2\pi}-1)(t/2\pi)\\[2ex]\mathsf E(X^2(t))&=(\mathrm e^{4\pi}-1)(t^2/4\pi)\\[2ex]\mathsf{Var}(X(t))&=(\mathrm e^{4\pi}-1)(t^2/4\pi)-(\mathrm e^{4\pi}-2\mathrm e^{2\pi}+1)(t^2/4\pi^2)\\[1ex]&=\big((\pi-1)\mathrm e^{4\pi}+2\mathrm e^{2\pi}-\pi-1\big)(t^2/4\pi^2)\end{align}$$