Ali Marshall

2022-04-27

I have to solve the following Cauchy's problem:

$\{\begin{array}{rl}& {x}^{2}{x}^{\prime}={\mathrm{sin}}^{2}({x}^{3}-3t)\\ & x(0)=1\end{array}.$

Addison Zamora

Beginner2022-04-28Added 10 answers

Step 1

$\{\begin{array}{c}{x}^{2}{x}^{\prime}={\mathrm{sin}}^{2}({x}^{3}-3t)\text{}\text{}......(1)\\ x(0)=1\end{array}.$

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

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

Now, (1) becomes,

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

$\Rightarrow \frac{dy}{dt}=-3\left({\mathrm{cos}}^{2}\left(y\right)\right)$

$\Rightarrow \frac{dy}{{\mathrm{cos}}^{2}\left(y\right)}=-3dt$

$\Rightarrow \mathrm{tan}\left(y\right)=-3t+c$

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

Using initial condition $x\left(0\right)=1$

$c=\mathrm{tan}\left(1\right)$

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

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