Laplace transform for solving differential equation ? help? I have a differential equation:
$$y'' + 4y' + 3y = 6t + 14, $$
with initial conditions $y(1) = 1,~y'(1) = 0.5$.
Can I use the Laplace transform to solve this equation whose initial condition is not at $0$ ?
 A: Of course that you can solve it in that way, just let $y(0) := c_1$ and $y'(0) := c_2$, as mentioned, and use your conditions at $t=1$ for solving them.
If you take the Laplace transform of your equation, you will have
$$
\begin{aligned}
\mathcal{L}_t\left[\underbrace{y''(t) + 4y'(t) + 3y(t) - 6t - 14}_{=0}\right](s) &= (s^2+4s+3)y(s) -(sy(0) + y'(0)) -4y(0) -\frac{6}{s^2} - \frac{14}{s} \\
&=(s^2+4s+3)y(s) - c_2 - (s+4)c_1 -\frac{6}{s^2} - \frac{14}{s}.
\end{aligned}
$$
Solving for $y(s)$, we have that
$$
y(s) = \frac{c_2 + (s+4)c_1 +6/s^2 + 14/s}{s^2+4s+3},
$$
and if we take the inverse Laplace transform for retrieving $y(t)$,
$$
\mathcal{L}^{-1}_s\left[y(s)\right](t) = \frac{1}{2}e^{-t}(3c_1+c_2-8) - \frac{1}{2}e^{-3t}(c_1+c_2-4) + 2t+2.
$$
Applying your conditions at $t=1$ in order to solve $c_1$ and $c_2$, we finally have that
$$
\boxed{y(t) = -\frac{21}{4}e^{1-t} + \frac{9}{4}e^{3(1-t)} +2t + 2}
$$
with
$$
c_1 = 2-\frac{21}{4}e+\frac{9}{4}e^3, \qquad c_2 = 2+\frac{21}{4}e -\frac{27}{4}e^3.
$$
Your solution will look like this:

As a final remark, notice that it would be much easier to solve this equation using the undetermined coefficients technique. Just solve the homogeneous equation using $y_h(t) = e^{\lambda t}$ and take as a particular solution $y_p(t) = c_1t+c_2$. Your general solution will be $y(t) = y_h(t) + y_p(t)$.
