The given equation and initial values are: $$\frac{d^2x}{dt^2}+25x=50e^{5t}$$ $$x(0)=0 \space, x^{'}(0)=0$$ Then taking the Laplace transform of the given: $$\mathscr{L}\left[x^{''}+25x \right]=\mathscr{L}[50e^{5t}]$$ $$s^2X(s)-x^{'}(0)-x(0)+25X(s)=\frac{50}{s-5}$$ Solving for $X(s)$: $$X(s)=\frac{50}{(s-5)(s^2+25)}$$ Then we have to decompose the above expression: $$\frac{A}{s-5}+\frac{Bs+c}{s^2+25}$$ $$50=A(s^2+25)+(Bs+C)(s-5)$$ Substituting in for $s=5$: $$50=50A \to A=1$$ Now we plug in our value for $A$ , then expand and factorizing yields: $$50=s^2(1+B)+s(-5B+C)+(-5C+25)$$ Now to equate the coefficients: $$(1):50=25-5C$$ $$(2):0=C-5B$$ $$(3):0=1+B$$ Solving equation $(1)$ for $C$ results with $C=-5$, then looking at equation $(3)$ we can deduce that $B=-1$
So now our Laplace transform looks like: $$X(s)=\frac{1}{s-5}+\frac{-s-5}{s^2+25}$$
After splitting up the second term and taking the inverse Laplace transform the answer I got is: $$e^{5t}-\cos(5t)-\sin(5t)$$
Your answer looks correct to me. Note that this IVP is simply: $$\frac{d^2x}{du^2}+x=2e^{u}$$ $$x(0)=0,x'(0)=0$$ Where $u=5t$. Apply Laplace's Transform: $$X(s)(s^2+1)=\dfrac 2 {s-1}$$ $$X(s)=\dfrac 2 {(s^2+1)(s-1)}$$ $$X(s)=\dfrac 1 {s-1}-\dfrac {(s+1)} {s^2+1}$$ Take inverse Lapalce transform: $$x(u)=e^u-\cos u -\sin u$$ $$\boxed {x(t)=e^{5t}-\cos (5t) -\sin (5t)}$$