Finding functional extremal with Euler-Lagrange equation

Question

I have $\int_1^2 (t\dot{x}^2+\frac{1}{t}x^2) dt$ with $x(1)=0$ and $x(2)=3$ and am trying to find the extremal. I know I need to start with the Euler-Lagrange equation $\frac{\partial f}{\partial x}-\frac{d}{dt} \frac{\partial f}{\partial \dot{x}}=0$ and have found $\frac{\partial f}{\partial x}=\frac{2}{t}x$ and that $\frac{\partial f}{\partial \dot{x}}=2t\dot{x}$ which leaves me with the EL equation $\frac{d}{dt}(t\dot{x})=\frac{1}{t}x$. However, at this point I get a bit stuck since the RHS of the equation depending on $x$ means I can't just integrate with respect to $t$. Does anyone have any tips on where to go from here?

Answer 1

Let $F(t,x,\dot x)=t\dot x^2+x^2/t$. As you have correctly stated, Euler-Lagrange yields $$\frac{dF_{\dot x}}{dt}=F_x\implies \frac d{dt}(2t\dot x)=\frac 2tx\implies \dot x+t\ddot x=\frac xt$$ using the Product Rule. The second-order ODE $$t^2\ddot x+t\dot x-x=0$$ is a simple Cauchy-Euler equation, from which you get $x=At+B/t$ for some constants $A,B$ that satisfy the boundary conditions.