Kelly Nelson

## Answered question

2021-12-28

Find a general solution of each of the following differential equations: $\frac{dy}{dx}=\frac{{x}^{6}{y}^{4}-2y}{x}$

### Answer & Explanation

macalpinee3

Beginner2021-12-29Added 29 answers

$⇒\frac{dy}{dx}+\frac{2}{x}y={x}^{5}{y}^{4}$
$⇒\frac{1}{{y}^{4}}\frac{dy}{dx}+\frac{2}{x}\frac{1}{{y}^{3}}={x}^{5}$
Let, $\frac{1}{{y}^{3}}=t⇒\frac{-3}{{y}^{4}}\frac{dy}{dx}=\frac{dt}{dx}$
$⇒\frac{-1}{dx}+\frac{2}{x}t={x}^{5}$
$⇒\frac{dt}{dx}-\frac{6}{x}t=-3{x}^{5}$
$I.F={e}^{\int \frac{-6}{x}dx}={e}^{-6\mathrm{ln}x}={e}^{\mathrm{ln}\left(\frac{1}{{x}^{6}}\right)}=\frac{1}{{x}^{6}}$
$t\left(\frac{1}{{x}^{6}}\right)=-\int 3{x}^{5}\frac{1}{{x}^{6}}dx+c$
$=-3\int \frac{1}{x}dx+c=-3\mathrm{ln}\left(x\right)+c$
$⇒t={x}^{6}\left(-3\mathrm{ln}\left(x\right)+c\right)$
$\therefore \frac{1}{{y}^{3}}=t$
$⇒\frac{1}{{y}^{3}}={x}^{6}\left(-3\mathrm{ln}\left(x\right)+c\right)$
$⇒y\left(x\right)=\frac{1}{\sqrt{3}\left\{{x}^{6}\left(-3\mathrm{ln}\left(x\right)+c\right)\right\}}$
or $y\left(x\right)=\frac{-\sqrt{3}\left\{-1\right\}}{\sqrt{3}\left\{{x}^{6}\left(-3\mathrm{ln}\left(x\right)+c\right)\right\}}$
or

Mary Herrera

Beginner2021-12-30Added 37 answers

$\frac{dy}{dx}=\frac{{x}^{6}{y}^{4}-2y}{x}$
$\frac{y}{x}=\frac{y\left(\frac{{x}^{6}{y}^{4}}{y}-\frac{2y}{y}\right)}{x}$
$\frac{y}{x}=\frac{y\left({x}^{6}{y}^{4-1}-2\right)}{x}$
$\frac{y}{x}=\frac{y\left({x}^{6}{y}^{3}-2\right)}{x}$
$y=y\left({x}^{6}{y}^{3}-2\right)$
$y=y{x}^{6}{y}^{3}-y\cdot 2$
$y=y{y}^{3}{x}^{6}-2y$
$y={y}^{1+3}{x}^{6}-2y$
$y={y}^{4}{x}^{6}-2y$
$y+\left(-{y}^{4}{x}^{6}+2y\right)=\left({y}^{4}{x}^{6}-2y\right)+\left(-{y}^{4}{x}^{6}+2y\right)$
$y-{y}^{4}{x}^{6}+2y={y}^{4}{x}^{6}-2y-{y}^{4}{x}^{6}+2y$
$y+2y-{y}^{4}{x}^{6}={y}^{4}{x}^{6}-{y}^{4}{x}^{6}-2y+2y$
$3y-{y}^{4}{x}^{6}=0$
$y\left(\frac{3y}{y}-\frac{{y}^{4}{x}^{6}}{y}\right)=0$
$y\left(3-\left({y}^{4-1}{x}^{6}\right)\right)=0$
$y\left(3-\left({y}^{3}{x}^{6}\right)\right)=0$
$y\left(-{y}^{3}{x}^{6}+3\right)=0$
$y=0$
$-{y}^{3}{x}^{6}+3=0$

karton

Expert2022-01-09Added 613 answers

Rewrite in form of a first order Bernoulli ODE
${y}^{\prime }+\frac{2}{x}y={x}^{5}{y}^{4}$
The general solution is obtained by substituting $v={y}^{1-n}$ and solving $\frac{1}{1-n}{v}^{\prime }+p\left(x\right)v=q\left(x\right)$
Transform to $\frac{1}{1-n}{v}^{\prime }+p\left(x\right)v=q\left(x\right):-\frac{1}{3}{v}^{\prime }+\frac{2v}{x}={x}^{5}$
Solve $-\frac{1}{3}{v}^{\prime }+\frac{2v}{x}={x}^{5}:v=-3\mathrm{ln}\left(x\right){x}^{6}+{c}_{1}{x}^{6}$
Substitute back $v={y}^{-3}:{y}^{-3}=-3\mathrm{ln}\left(x\right){x}^{6}+{c}_{1}{x}^{6}$
Isolate y: $y=\sqrt[3]{\frac{1}{{x}^{6}\left(-3\mathrm{ln}\left(x\right)+{c}_{1}\right)}}$
$y=\sqrt[3]{{x}^{6}\left(-3\mathrm{ln}\left(x\right)+{c}_{1}\right)}$

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