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I have to solve the following Cauchy's problem:

$$ \left\{ \begin{align} & x^2 x'=\sin^2(x^3-3t) \\ & x(0)=1 \end{align} \right. $$


First of all I've tried to identify if it is homogeneous, linear, exact, euler's ... but I can't recognise it. So I don't know which procedure to follow to solve it...

I think that my problem is that inside the $\sin$ there are both $x$ and $t$.

Could anyone help me?

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    $\begingroup$ Maybe letting $y(t) = x^3(t)$ might be actually helpful, because $y'(t) = 3x^2(t)x'(t)$ so you get equivalent equation $y(0)=1, y' = 3\sin^2(y-t)$. Edit: Then letting $z(t) = y(t)-t$ you get $z'(t) = y'(t) - 1$, so that $z' = 3\sin^2(z) - 1$. This should be solvable $\endgroup$ Commented Jan 6, 2022 at 21:00
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    $\begingroup$ Somehow I've missed $3$ in front of $t$, and unfortunatelly I can't edit my comment anymore. Obviously the substitution should be $z=x^3-3t$, which will give the equation $z' = 3\sin^2(z) - 3$ with $z(0)=1$. $\endgroup$ Commented Jan 6, 2022 at 21:18
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    $\begingroup$ Thanks! I have done it as you've said but in the end I've got a doubt: I obtain $\frac{-1}{3}\tan(x^3 -3t)=t+K$, being $K$ a constant. But what do I have to do with $x(0)=1$? I have to put $0$ where there is $t$ and reach to $\frac{-1}{3}\tan(x^3)=K$??? @DominikKutek $\endgroup$
    – User160
    Commented Jan 6, 2022 at 21:24
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    $\begingroup$ Yes, since you've get equation $\tan(x^3 - 3t) = -3t + C$ for some constant $C \in \mathbb R$, you simply let $t=0$ (so that $x(0)=1$), so $\tan(x^3(0) - 3\cdot 0) = -3 \cdot 0 + C$ meaning $C = \tan(1)$. $\endgroup$ Commented Jan 6, 2022 at 22:38

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$\left\{\begin{matrix} x^2 x'=\sin^2(x^3-3t) \space \space ......(1)\\ x(0)=1 \end{matrix}\right.$

Let, $x^3-3t=y$

$3x^2\frac{dx}{dt}-3=\frac{dy}{dt}$

Now, (1) becomes,

$\frac{dy}{dt}=3( \sin^2(y)-1)$

$\implies \frac{dy}{dt}=-3(\cos^2(y))$

$\implies \frac{dy}{cos^2(y) }=-3dt$

$\implies \tan(y) =-3t+c$

Hence, $\tan(x^3-3t) =-3t+c$

Using initial condition $x(0) =1$

$c=\tan(1) $

Hence, $x=\sqrt[3]{3t+\tan^{-1}(tan(1) -3t) }$

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  • $\begingroup$ Why can't you keep the variables as they were? Or if you want to change their names, at least don't use the same letters as in the problem. $\endgroup$
    – MasB
    Commented Jan 6, 2022 at 22:15
  • $\begingroup$ Ok. I have understood the problem, thanks $\endgroup$ Commented Jan 7, 2022 at 4:16

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