Let the two coupled Euler-Lagrange equations giving the stationary path of the functional $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx$ be $y_{1}''-2(2y_{1}+y_{2})=0$ and $2y_{2}''-(2y_{1}+y_{2})=0$. By making the linear transformation $z_{1}=y_{1}+ay_{2}, z_{2}=2y_{1}+y_{2}$ to the new dependent variables $(z_{1}, z_{2})$, where $a$ is a constant, the value of $a$ such that the functional can be written as the sum of two functionals, one of which depends only on $z_{1}$ and the other only on $z_{2}$ is $-4$.
a) Find the general solution of the Euler-Lagrange equations for $(z_{1}, z_{2})$, and hence find the general solution for $(y_{1}, y_{2})$.
b) Verify by direct substitution that your general solution for $(y_{1}, y_{2})$ satisfies the two coupled Euler-Lagrange equations giving the stationary path of the functional $S[y_{1}, y_{2}]=\int [y_{1}'^2+2y_{2}'^2+(2y_{1}+y_{2})^2]dx$.
I know that the Euler-Lagrange equation is $\frac{d}{dx}(\frac{\partial F}{\partial y'})-\frac{\partial F}{\partial y}=0, y(a)=A, y(b)=B$ for the functional $S[y]=\int_{a}^{b}F(x, y, y')dx, y(a)=A, y(b)=B$, but how should I find the Euler-Lagrange equations for $z_{1}=y_{1}+ay_{2}, z_{2}=2y_{1}+y_{2}$?
By considering the functional $S[z_{1}, z_{2}]=\frac{1}{9}(\int_{x_{0}}^{x_{1}}z_{1}'^2dx+\int_{x_{0}}^{x_{1}}(2z_{2}'^2+9z_{2}^2)dx)$ and applying the definition of the Euler-Lagrange equation, we have that the Euler-Lagrange euqation for $z_{1}$ is $\frac{2}{9}z_{1}''=0$ and the Euler-Lagrange equation for $z_{2}$ is $\frac{4}{9}z_{2}''-2z_{2}=0$.
From here after integrating, we can see that the general solution of the Euler-Lagrange equation for $z_{1}$ is $z_{1}=\frac{9}{2}(c_{1}x+c_{2})$ and the general solution of the Euler-Lagrange equation for $z_{2}$ is $z_{2}=c_{3}e^{\frac{3x}{\sqrt{2}}}+c_{4}e^{-\frac{3x}{\sqrt{2}}}$.
Note that the Euler-Lagrange equation for $y_{1}$ is $y_{1}''-2(2y_{1}+y_{2})=0$ and the Euler-Lagrange equation for $y_{2}$ is $2y_{2}''-(2y_{1}+y_{2})=0$.
The general solution for $(y_{1}, y_{2})$ is $y_{1}=\frac{c_{1}x+c_{2}+4(c_{3}e^{\frac{3x}{\sqrt{2}}}+c_{4}e^{-\frac{3x}{\sqrt{2}}})}{9}$ and $y_{2}=\frac{-2(c_{1}x+c_{2})+c_{3}e^{\frac{3x}{\sqrt{2}}}+c_{4}e^{-\frac{3x}{\sqrt{2}}}}{9}$.
By direct substitution, we have $0=0$.