$y'' -4y' = -4te^{2t}, y(0)=0, y'(0)=1$ If you take laplace Transform of all terms, isolate L(y), I got $$L(y) = \frac{1}{(p-2)^2} + \frac{-2}{(p-2)^2 -4}$$
Then, taking inverse Laplace, you get $$y(t) = te^{2t} - e^{2t}sinh(2t)$$
But the solution is just $te^{2t}$. WHat am I doing wrong
On
Your solution works for the initial conditions $y(0)=0,y'(0)=-1$, not the ones given in the question.
$\mathcal L[y'']-4\mathcal L[y']=[s^2Y(s)-sy(0)-y'(0)]-4[sY(s)-y(0)]\\=(s^2-4s)Y(s)\color{red}{-1}=\dfrac{-4}{(s-2)^2}\\\therefore Y(s)=\dfrac1{s^2-4s}-\dfrac4{(s^2-4s)(s-2)^2}=\dfrac1{(s-2)^2}\\\therefore y(t)=te^{2t}$
On
First you should determine whether you want to use one or two sided Laplace transform. This problem can be solved easily using two sided Laplace transform in which case the initial conditions should not be applied. By definition$$F(s)=\int_{-\infty}^{\infty}f(t)e^{-st}dt$$and$$f'(t)\iff sF(s)\\f''(t)\iff s^2F(s)$$also $$e^{at}\iff \delta(s-a)$$therefore $$te^{at}\iff -\delta'(s-a)$$by substitution we obtain $$(s^2-4s)Y(s)=-4\delta'(s-2)\iff Y(s)={-4\delta'(s-2)\over s^2-4s}=-\delta'(s-2)$$therefore $$y(t)=te^{2t}$$
What are the initial conditions? What is the given solution?
Your $-e^{2t}\sinh(2t)$ is a solution to the homogeneous equation $y''-4y'=0$, and can also be written as $\frac12e^{4t}-\frac12$. The homogeneous equation has solutions $ae^{4t}+b$ for constants $a$ and $b$. If there are no initial conditions given, we take all of them, while if there are specific initial conditions we'll need specific values - which can be found by applying appropriate care in the Laplace transform equation.
Given initial conditions $y(0)=c_0$, $y'(0)=c_1$, the Laplace transform equation becomes $$(p^2L(p) - pc_0 - c_1) - 4(pL(p) - c_0) = \frac{-4}{(p-2)^2}$$ Solve that for $L(p)$, and you'll have the exact solution with initial conditions. If your form has hyperbolic functions $\cosh$ or $\sinh$ in it, remember that those can be converted into exponentials.