I have to solve the following Cauchy's problem:{x2x′=sin2⁡(x3−3t)x(0)=1.

Step 1{x2x′=sin2⁡(x3−3t)&nbsp;&nbsp;......(1)x(0)=1.Let,&nbsp;x3−3t=y3x2&nbsp;dx&nbsp;&nbsp;dt&nbsp;−3=&nbsp;dy&nbsp;&nbsp;dt&nbsp;Now, (1) becomes,&nbsp;dy&nbsp;&nbsp;dt&nbsp;=3(sin2(y)−1)⇒&nbsp;dy&nbsp;&nbsp;dt&nbsp;=−3(cos2(y))⇒&nbsp;dy&nbsp;cos2(y)=−3&nbsp;dt&nbsp;⇒tan⁡(y)=−3t+cHence,&nbsp;tan⁡(x3−3t)=−3t+cUsing initial condition&nbsp;x(0)=1c=tan⁡(1)Hence,&nbsp;x=3t+tan−1(tan⁡(1)−3t)3

Solved: "I have to solve the following Cauchy's problem:{x2x′=sin2⁡(x3−3t)x(0)=1."

High School
Calculus and Analysis
Calculus 2
Differential equations

Ali Marshall

Answered question

2022-04-27

I have to solve the following Cauchy's problem:

{\begin{aligned} x^{2} x^{'} = \sin^{2} (x^{3} - 3 t) \\ x (0) = 1 \end{aligned} .

Answer & Explanation

Addison Zamora

Beginner2022-04-28Added 10 answers

Step 1
${\begin{matrix} x^{2} x^{'} = \sin^{2} (x^{3} - 3 t) . . . . . . (1) \\ x (0) = 1 \end{matrix} .$
Let, $x^{3} - 3 t = y$
$3 x^{2} \frac{d x}{d t} - 3 = \frac{d y}{d t}$
Now, (1) becomes,
$\frac{d y}{d t} = 3 (\sin^{2} (y) - 1)$
$\Rightarrow \frac{d y}{d t} = - 3 (\cos^{2} (y))$
$\Rightarrow \frac{d y}{\cos^{2} (y)} = - 3 d t$
$\Rightarrow \tan (y) = - 3 t + c$
Hence, $\tan (x^{3} - 3 t) = - 3 t + c$
Using initial condition $x (0) = 1$
$c = \tan (1)$
Hence, $x = \sqrt[3]{3 t + \tan^{- 1} (\tan (1) - 3 t)}$

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