Find the solution to the first order ODE's Find the solution to 
$x′= y−x+t $
and
$y'= y $
if $x(0)=8$ and $y(0)=2$
I'm confused how to go about this question. If anyone could offer some clarification that would be greatly appreciated!
 A: First solve the second equation as it does not contain $x$. Then insert
$$
y(t)=2e^t
$$
into the first equation and solve via your preferred method, integrating factor, homogeneous plus particular solution etc.
A: HINT
The system is in the form
$$
\begin{bmatrix}
x^\prime \\y^\prime
\end{bmatrix} = 
\begin{bmatrix}
-1 & 1\\ 0 & 1
\end{bmatrix}
\begin{bmatrix}
x \\y
\end{bmatrix}
+t\begin{bmatrix}
1 \\0
\end{bmatrix} \quad \vec x^{\prime}=A \vec x+t\vec b$$
Firstly we need to find the complementary solution to $\vec x^{\prime}=A \vec x$ assuming $\vec x=\vec u\cdot e^{at}$ that is
$$a\cdot \vec u\cdot e^{at}=A\vec u\cdot e^{at}\implies(A-aI)\vec u=0$$
and from here we can find the eigenvalues, the eigenvectors and the complementary solution $\vec x_c$ to the system.
Then we need to find the particular solution to $\vec x^{\prime}=A \vec x+t\vec b$ assuming 
$$\vec x_p=t\vec v + \vec w \implies \vec x_p^{\prime}=\vec v$$
and then
$$\vec v=tA\vec v+A\vec w+t\vec b\implies t(A\vec v+\vec b)+(A\vec w-\vec u)=0 $$
that is


*

*$A\vec v+\vec b=\vec 0$ from which we find $\vec v$

*$A\vec w-\vec v=\vec 0$ from which we find $\vec w$


Then the general solution is $\vec x=\vec x_c+\vec x_p$ and we can impose the initial condition to find the constants in $\vec x_c$.
Here you can find some worked example.
A: First solve the second equation 
$$y'=y \implies \ln |y|=t+K_1 \implies y=K_1e^t$$
Then solve first equation
$$x′= y−x+t$$
$$x'+x=K_1e^t+t$$
Multiply by $e^t$
$$(xe^t)'=K_1e^{2t}+te^t$$
Integrate
$$xe^t= \int K_1e^{2t}+te^t dt $$
$$......$$
