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Can you help me to solve this DF:
$tx''+\frac{2}{3} x'=0$ with initial conditions $x(0)=0$, $x(1)=1$, $x=x(t)$

I've done this:
$tx''+\frac{2}{3} x'=0$
$x''+\frac{2}{3t} x'=0$
$r^2+\frac{2}{3t}r=0$
$r(r+\frac{2}{3t})=0$
$r_1=0$, $r_2=-\frac{2}{3t}$
$x=C_1e^{r_1 t}+C_2e^{r_2 t}=C_1e^{0t}+C_2e^{-\frac{2}{3t}t}=C_1+C_2e^{-\frac{2}{3}}$ which probably can't be good because I can't apply initial conditions on this.

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Using the integrating factor method, we find that

$$(t^{2/3}x')'=t^{2/3}x''+\tfrac{2}{3}t^{-1/3}x'=t^{-1/3}(tx''+\tfrac{2}{3}x')=0.$$

It follows that $t^{2/3}x'=C_1$, so $x=3C_1t^{1/3}+C_2$. Now, $x(0)=0$ gives $C_2=0$ and $x(1)=1$ gives $C_1=\frac{1}{3}$. Hence the solution to the boundary value problem is given by $x(t)=\sqrt[3]{t}$.

Your method does not work because the coefficients are not constant.

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You need to make sure you're applying the right rules to the right questions- the one you've tried to do only works when the coefficients of $x, x'$ and $x''$ are constants.

Hint: Substitute in $y=x'$

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