Find the solution $\Phi$ of an IVP Find the solution $\Phi$ of the system
\begin{align*}
y_1'&=-y_1\\
y_2'&=y_1+ty_2\\
\end{align*}
satisfying the initial condition $\Phi(0)=(2,1)$.
It's been years since I've taken ODE, so I don't remember how to do these. I solved a similar problem, pretty much the same one but without the "t" variable in front of $y_2$ (the "t" is throwing me off). For $y_1$ I got $$y_1=2e^{-t}$$ 
I need help as to how to solve $y_2$. 
 A: You solved the problem for the first equation and obtained $y_1=2e^t$. As said by Amzoti, substitute $y_1$ epression in the second equation and you arrive to $$y_2' =2 e^{-t}+t y_2$$ So, as usual, you first solve  $$y_2' =t y_2$$ which leads to $$y_2=K(t) e^{\frac{t^2}{2}}$$ which you differentiate again and bring back to the differential equation. This now leads to $$e^{\frac{t^2}{2}} K'(t)=2 e^{-t}$$ that is to say $$K'(t)=2 e^{-\frac{t^2}{2}-t}$$ which you need to integrate. As Amzoti said, this is not the most pleasant integration but a rather simple change of variable leads to the solution $$K(t)=\sqrt{2 e \pi } \text{erf}\left(\frac{t+1}{\sqrt{2}}\right)+C$$ so the solution is $$y_2=e^{\frac{t^2}{2}} \left(C+\sqrt{2 e \pi }
   \text{erf}\left(\frac{t+1}{\sqrt{2}}\right)\right)$$ Applying the boundary condition, you obtain  $$C=1-\sqrt{2 e \pi } \text{erf}\left(\frac{1}{\sqrt{2}}\right)$$ and so $$y_2=e^{\frac{t^2}{2}} \left(\sqrt{2 e \pi }
   \left(\text{erf}\left(\frac{t+1}{\sqrt{2}}\right)-\text{erf}\left(\frac{1}{\sqrt{2
   }}\right)\right)+1\right)$$ A tedious differentiation will confirm that the second differential equation and its boundary condition are satisfied with this expression.  
In this answer, I tried to go into as much details to refresh your memory. Do not hesitate to post if you need further clarification.
