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Can someone please give me a hint on this problem? I don't know how to write Euler equation for this case:

Find the extremal for the functional $$ J(x)=\int_1^{t_f} \dot{x}^2(t)t^3\,dt $$ which has $x(1)=0$ and $x(t_f)$ must lie on $\theta(t)=\dfrac{2}{t^2}-3$

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I would go back to first principles and find the Gâteaux derivative $\Delta J[x,\psi]$ of $J$. This has to be zero for $\psi$ to be a stationary path.

With $G=x'^2t^3$, $G(x+\tau\psi)=(x'+\tau\psi')^2t^3$, and $$\frac{d}{d\tau} G(x+\tau\psi) = 2(x'+\tau\psi')\psi't^3$$ $$\left.\frac{d}{d\tau} G(x+\tau\psi)\right|_{\tau=0}= 2x'\psi't^3$$

Plug this into $J$ to get an expression for $\Delta J$. $$ \Delta J[x,\psi] = \left.\frac{d}{d\tau} J(x+\tau\psi)\right|_{\tau=0}$$

Then use integration by parts. This gives two terms. The first term, the $[\cdot]$ term, can be simplified because at the lower boundary $\psi(1)=0$. For second, $\int\cdot dt$, term, note that some $\psi$ are zero at the upper boundary (in which case the first term disappears and the second term must be zero), so you can use the fundamental lemma here. As a result the second term has to be zero, and so the first term also has to be zero. Putting both together should give you a differential equation for $x$.

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