Question

solve Y" -3y' + 2y = 4t+e^3t , y(0)=1, y'(0)=1 by laplace inverse transform

Answer 1

characteristic equation:
\[r^2 - 3r+2 = 0\]
\[r = 1,2\]
Homogeneous solution:
\[y_c (t) = c_1 e^{t} + c_2 e^{2t}\]
Particular solution:
\[y_p (t) = At+B e^{3t}+C\]
\[\rightarrow 9B e^{3t} -3(A+3Be^{3t})+2(At +Be^{3t} +C) = 4t+e^{3t}\]
\[2A = 4 \rightarrow A = 2\]
\[2B = 1 \rightarrow B = \frac{1}{2}\]
\[-3A +2C = 0 \rightarrow C = 3\]
General Solution:
\[y(t) = c_1 e^t +c_2 e^{2t} + 2t + \frac{1}{2} e^{3t} + 3\]
Apply initial conditions
\[c_1 + c_2 + \frac{1}{2} = 1\]
\[c_1 + 2c_2 + \frac{3}{2} = 1\]
\[c_1 = \frac{3}{2} , c_2 = -1\]
Final Solution:
\[Y(t) = \frac{3}{2}e^t -e^{2t} +2t + \frac{1}{2}e^{3t} +3\]

Answer 2

Ingredients:
\[\begin{array}{c|c}
f(t)\text{ (original function)}&F(s)\text{ (transform of original function)}\\[1ex]
\hline
f^{(n)}(t)&s^nF(s)-\sum\limits_{k=1}^n s^{n-k}f^{(k-1)}(0)\\[1ex]
\hline
x^n&\dfrac{n!}{s^{n+1}}\\[1ex]
\hline
e^{kt}&\dfrac{1}{s-k}
\end{array}\]where \(f^{(n)}\) denotes the \(n\)th derivative of \(f\), and in particular you have
\[\mathcal{L}\left\{f''(t)\right\}=s^2F(s)-sf(0)-f'(0)\]and
\[\mathcal{L}\left\{f'(t)\right\}=sF(s)-f(0)\]
Taking the transform of both sides, we thus go from
\[y''-3y'+2y=4t+e^{3t}\]to
\[\begin{align*}\Bigg[s^2F(s)-sy(0)-y'(0)\Bigg]-3\Bigg[sF(s)-y(0)\Bigg]+2F(s)&=\frac{4}{s^2}+\frac{1}{s-3}\\[1ex]
F(s)\left(s^2-3s+2\right)-s+2&=\frac{4}{s^2}+\frac{1}{s-3}\\[1ex]
F(s)&=\frac{s^3-3s^2+s+6}{s^2(s-1)(s-3)}\end{align*}\]From here, you're going to want to split up the right side into partial fractions and make finding the inverse transform (and hence your solution) much easier.