I'm not really up on "control" problems and I have no idea what being "cost functional" means but this looks to me like solving a system of differential equations. I am going to write the independent variable as "t".
If you are used to dealing with "control" problems, you might be used to setting a system of differential equations up as a matrix equation. For this problem that would be $\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}'= \begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$$\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$$+ \begin{pmatrix} 0 \\ u \end{pmatrix}$
and then observe that the matrix $\begin{pmatrix}1 & 1 \\ 0 & 1 \end{pmatrix}$ has 1 as a double eigenvalue.
Personally I wouldn't get that "sophisticated". I would, rather, note that the second equation, $y_2'= y_2+ u$ doesn't involve $y_1$ so can be solved directly. The "associated homogeneous equation" is $y_2'= y_2$ which has the general solution $y_2= Ce^t$ for C any constant. Using the "variation of parameters" method, we seek a solution of the form $y_2= v(t)e^t$ (thus "varying the parameter"- letting C be a function). $y_2'= v'e^t+ ve^t= ve^t+ u$ so that $v'e^t= u$, $v'= e^{-t}u$ and $v= \int_1^t e^{-z}u(z)dz$ (I have taken 1 as the lower limit in order to simplify using $y_2(1)= 0$).
So $y_2(t)= Ce^t+ e^t\int_1^t e^{-z}u(z)dz$. Setting t= 1, $y_2(1)= Ce= 0$ so $C= 0$. $y_2(t)= e^t\int_1^t e^{-z}u(z)dz$.
Now turn to the first equation. $y_1'= y_1+ y_2= y_1+ e^t\int_1^t e^{-z}u(z)dz$. The "associated homogeneous equation" for this is $y_1'= y_1$ which has general solution $y_1(t)= De^t$. Again, to use "variation of parameters" we seek a solution of the form $y_1(t)= w(t)e^t$. $y_1'= w'e^t+ we^t= we^t+ e^t\int_1^t e^{-z}u(z)dz$ so $w' e^t= e^t\int_1^t e^{-z}u(z)dz$. $w'(t)= \int_1^z e^{-z}u(z)dz$ and $w(t)= \int_1^t\int_1^s e^{-z}u(z)dz ds$.
Then $y_1(t)= De^t+ e^t\int_1^t\int_1^s e^{-z}u(z)dz ds$.
Since $y_1(1)= D= 0$, $y_1(t)= e^t\int_1^t\int_1^s e^{-z}u(z)dz ds$.
Now that we have solved the differential equations in terms of "u" what would make u a "control"?
