How can I solve the following exercise

Question

Find the critical curves for the following functional : $$J[y(x),z(x)]=\int_{0}^{1}(y'^2+z'^2-xyz'-yz)dx$$ With the conditions : $$K[y(x),z(x)]=\int_{0}^{1}(y'^2-xy'-z'^2)dx=2$$ $$y(1)=z(1)=1$$ $$y(0)=z(0)=0$$ We know that is functional with subsidiary conditions So first we define $$H(x,y,z,y',z',λ(x))=F+λ(x)G$$ . Such that $$F(x,y,z,y',z')=y'^2+z'^2-xyz'-yz$$ and $$G(x,y,z,y',z')=y'^2-xy'-z'^2$$ $$H(x,y,z,y',z',λ(x))=y'^2+z'^2-xyz'-yz+λ(x)(y'^2-xy'-z'^2)$$ So we have to find Euler -Lagrange equations : $$H_y-\frac{dH_y'}{dx}=0$$ and $$H_z-\frac{dH_z'}{dx}=0$$. $H_y$ refers to the partial derivative of $H$ With respect to $y$ and the same for others . $$H_y=-xz'-z$$ and $$H_y'=2y'+2λ(x)y'-xλ(x)$$ then $$\frac{dH_y'}{dx}=2y''+2λ'(x)y'+2λ(x)y''-λ(x)-xλ'(x)$$ So the first equation is $$-xz'-z-2y''-2λ'(x)y'-2λ(x)y''+λ(x)+xλ'(x)=0$$ $$H_z=-y$$ and $$H_z'=2z'-xy-2λ(x)z'$$ then $$\frac{dH_z'}{dx}=2z''-y-xy'-2λ'(x)z'-2λ(x)z''$$ so the second equation is $$-y-2z''+y+xy'+2λ'(x)z'+2λ(x)z''=0$$ so $$-2z''+xy'+2λ'(x)z'+2λ(x)z''=0$$ after that I couldn't complete , I don't know how can I solve like these differential equations I mean the first and the second equations to find the critical curves for the following functional , so I want some help from you , please

Answer 1

Since your condition is an integral condition, and not pointwise along the path, the Lagrange multiplier $λ$ is a constant, which should somehow reduce the equations.

In the end, you obtain solution curves that depend on $λ$, as does in consequence the value of the condition integral. Now you have a one-parameter problem $h(λ)=2$ to solve, where the internals of $h(λ)$ contain all the integration tasks.