Checking Initial Values of an IVP Show that the solution of the initial value problem
$$y''+p(t)y'+q(t)y=g(t),y(t_0)=y_0, y'(t_0)=y_0'$$
can be written as $y=u(t)+v(t)$, where $u$ and $v$ are solutions of the two initial value problems
$$u''+p(t)u'+q(t)u=0, u(t_0)=y_0, u'(t_0)=y_0'$$
$$v''+p(t)v'+q(t)v=g(t), v(t_0)=0, v'(t_0)=0$$
respectively. Be sure to check the initial values.
I was able to show that the solution to the IVP can be written as $y=u(t)+v(t)$ (I believe).
$$(u+v)''+p(t)(u+v)'+q(u+v)=[u''(t)+v''(t)]+[p(t)(u'(t)+v'(t))]+[q(t)(u(t)+v(t))]$$
$$=(u''+pu'+qu)+(v''+pv'+qv)$$
$$=0+g(t)$$
$$=g(t)$$
The problem I am having is checking the initial values. Generally, when it's a simple
$$y''+y'-2y=2t, y(0)=0, y'(0)=1$$
type of problem, checking the initial values is very straightforward. For some reason or another, I can't quite seem to check the initial values on this one. I'm sure it is something silly that I'm having a mental block on because it looks different than normal.
Note: this is problem #21 in section 3.7 from Elementary Differential Equations and Boundary Value Problems, Eighth Edition. Boyce, DiPrima.

For the u part: y''+y'-2y=0, y(0)=0,y'(0)=1 
Solve the characteristic polynomial, you can find the general solution to be $u=C_1e^{-2t}+C_2e^t$. Put the initial value in, that is $C_1+C_2=0$ and $-2C_1+C2=1$. Subtract the first one from the second one you can find $C_1=-\frac{1}{3}$ then $C_2=\frac{1}{3}$
For the v part: y''+y'-2y=2t, y(0)=0,y'(0)=0  
The general solution is $v=C_3e^{-2t}+C_4e^t-t-\frac{1}{2}$. Put the initial value in, that is $C_3+C_4=\frac{1}{2}$ and $-2C_3+C_4=1$. Subtract the first one from the second one you can find $C_3=-\frac{1}{6}$ then $C_4=\frac{2}{3}$
So $u=-\frac{1}{3}e^{-2t}+\frac{1}{3}e^t$ and $v=-\frac{1}{6}e^{-2t}+\frac{2}{3}e^t-t-\frac{1}{2}$ Add them together we have $y=-\frac{1}{2}e^{-2t}+e^t-t-\frac{1}{2}$
For this case $y(0)=-\frac{1}{2}+1-0-\frac{1}{2}=0$ and $y'(0)=-\frac{1}{2}*(-2)+1-0-\frac{1}{2}=1$ 
Is this what you want?
