Find a general solution and then particular solution satisfying $y(0) = 0$ and $y'(0) = 1$ for $y'' – y = e^{2x}$
I have found the general solution when the ODE equals 0 to be $y = c_1e^x + c_2e^{-x} + \frac{1}{3}e^{2x}$, where the undetermined coefficient is $\frac{1}{3}$.
I am having trouble using the given initial values to get the constants $c_1$ and $c_2$ though and not sure where to start. My only observation at this point is that since $y' = 1$, then $y'' = 0$ I think.
Best Answer
Since you get $$y=c_1e^x+c_2e^{-x}+\frac{1}{3}e^{2x}$$
Calculate $$y'=c_1e^x-c_2e^{-x}+\frac{2}{3}e^{2x}$$
At $x=0$, $$y|_{x=0}=c_1+c_2+\frac{1}{3}=0 \implies c_1+c_2=-\frac{1}{3}$$
$$y'|_{x=0}=c_1-c_2+\frac{2}{3}=1 \implies c_1-c_2=\frac{1}{3}$$
Now simultaneous equation and solve for the coefficients :)