Find the two coupled Euler-Lagrange equations giving the stationary path of the functional
$$S[y_{1}, y_{2}]=\int \left[y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2\right]dx$$
Here's my work:
Consider the functional $$S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx$$
Let $L=y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2$.
Now we consider two cases of $y_{1}$ and $y_{2}$.
Case #1: Consider the function $y_{1}$.
Let the Euler-Lagrange equation be of the form $\frac{d}{dx}(\frac{\partial L}{\partial y_{1}'})-\frac{\partial L}{\partial y_{1}}=0$.
Note that $\frac{\partial L}{\partial y_{1}'}=2y_{1}'$ and $\frac{\partial L}{\partial y_{1}}=4(2y_{1}+y_{2})$.
Then $\frac{d}{dx}(\frac{\partial L}{\partial y_{1}'})=2y_{1}''$.
Observe that $\frac{d}{dx}(\frac{\partial L}{\partial y_{1}'})-\frac{\partial L}{\partial y_{1}}=0\implies \\ 2y_{1}''-4(2y_{1}+y_{2})=0$
Thus, the Euler-Lagrange equation for the function $y_{1}$ is $y_{1}''-2(2y_{1}+y_{2})=0$.
Case #2: Consider the function $y_{2}$.
Let the Euler-Lagrange equation be of the form $\frac{d}{dx}(\frac{\partial L}{\partial y_{2}'})-\frac{\partial L}{\partial y_{2}}=0$.
Note that $\frac{\partial L}{\partial y_{2}'}=4y_{2}'$ and $\frac{\partial L}{\partial y_{2}}=2(2y_{1}+y_{2})$.
Then $\frac{d}{dx}(\frac{\partial L}{\partial y_{2}'})=4y_{2}''$.
Observe that $\frac{d}{dx}(\frac{\partial L}{\partial y_{2}'})-\frac{\partial L}{\partial y_{2}}=0\implies 4y_{2}''-2(2y_{1}+y_{2})=0$.
Thus, the Euler-Lagrange equation for the function $y_{2}$ is $2y_{2}''-(2y_{1}+y_{2})=0$.
Therefore, the two coupled Euler-Lagrange equations giving the stationary path of the functional $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]$ are $y_{1}''-2(2y_{1}+y_{2})=0$ and $2y_{2}''-(2y_{1}+y_{2})=0$.
I want to know if the above work with the answer is correct or not. Please check/verify/review them and let me know if something is incorrect.