piarepm

2021-12-26

$\left(3x{y}^{2}-9xy-30x\right)dx+\left({x}^{2}+9\right)dy=0$

Jacob Homer

Step 1
Separate the variables both sides then integrate to get the answer.
Step 2
Given. $\left(3x{y}^{2}-9xy-30x\right)dx+\left({x}^{2}+9\right)dy=0$
$⇒3x\left({y}^{2}-3y-10\right)dx+\left({x}^{2}+9\right)\cdot dy=0$
$⇒3x\left({y}^{2}+2y-5y-10\right)dx+\left({x}^{2}+9\right)dy=0$
$⇒3x\left(y\left(y+2\right)-5\left(y+2\right)\right)dx+\left({x}^{2}+9\right)dy=0$
$⇒3x\left(y+2\right)\left(y-5\right)dx+\left({x}^{2}+9\right)dy=0$
devide the whole equation by $\left(y+2\right)\left(y-5\right)\left({x}^{2}+9\right)$
$⇒\frac{3x}{{x}^{2}+9}dx+\frac{1}{\left(y+2\right)\left(y-5\right)}\cdot dy=0$
$⇒\frac{1}{\left(y+2\right)\left(y-5\right)}dx=-\frac{3x}{{x}^{2}+9}dx$
$\frac{\left(y+2\right)-\left(y-5\right)}{\left(y+2\right)\left(y-5\right)}dy=-\frac{3}{2}\cdot \frac{2x}{{x}^{2}+9}dx$
$⇒\frac{1}{7}\left(\frac{1}{y-5}-\frac{1}{y+2}\right)dy=-\frac{3}{2}\frac{2x}{{x}^{2}+9}dx$
Integrating both sides

Piosellisf

$\left(3x{y}^{2}-9xy-30x\right)dx+\left({x}^{2}+9\right)dy=0$
$3x\left({y}^{2}-3y-10\right)dx+\left({x}^{2}+9\right)dy=0$
$3x\left({y}^{2}-5y+2y-10\right)dx=-\left({x}^{2}+9\right)dy$
$\frac{3xdx}{{x}^{2}+9}=-\frac{dy}{\left({y}^{2}-5y+2y-10\right)}$
$\frac{3}{2}\cdot \frac{2xdx}{{x}^{2}+9}=\frac{-dy}{y\left(y-5\right)+2\left(y-5\right)}$
$\frac{3}{2}\cdot \frac{2xdx}{{x}^{2}+9}=\frac{-dy}{\left(y+2\right)\left(y-5\right)}$
$\frac{3}{2}\cdot \frac{2xdx}{{x}^{2}+9}=\frac{1}{7}\left(\frac{1}{y+2}-\frac{1}{y-5}\right)dy$
Integrat both side
$\frac{3}{2}\int \frac{2x}{{x}^{2}+9}dx=\frac{1}{7}\int \left(\frac{1}{y+2}-\frac{1}{y-5}\right)dy$
$\frac{3}{2}\mathrm{ln}\left({x}^{2}+9\right)=\frac{1}{7}\left[\mathrm{ln}\left(y+2\right)-\mathrm{ln}\left(y-5\right)\right]$
$\frac{3}{2}\mathrm{ln}\left({x}^{2}+9\right)=\frac{1}{7}\mathrm{ln}\left[\frac{y+2}{y-5}\right]+c$

user_27qwe

$\begin{array}{}⇒3x\left({y}^{2}-3y-10\right)dx+\left({x}^{2}+9\right)\cdot dy=0\\ ⇒3x\left({y}^{2}+2y-5y-10\right)dx+\left({x}^{2}+9\right)dy=0\\ ⇒3x\left(y\left(y+2\right)-5\left(y+2\right)\right)dx+\left({x}^{2}+9\right)dy=0\\ ⇒3x\left(y+2\right)\left(y-5\right)dx+\left({x}^{2}+9\right)dy=0\\ \text{devide the whole equation by}\left(y+2\right)\left(y-5\right)\left({x}^{2}+9\right)\\ ⇒\frac{3x}{{x}^{2}+9}dx+\frac{1}{\left(y+2\right)\left(y-5\right)}\cdot dy=0\\ ⇒\frac{1}{\left(y+2\right)\left(y-5\right)}dx=-\frac{3x}{{x}^{2}+9}dx\\ \frac{\left(y+2\right)-\left(y-5\right)}{\left(y+2\right)\left(y-5\right)}dy=-\frac{3}{2}\cdot \frac{2x}{{x}^{2}+9}dx\\ ⇒\frac{1}{7}\left(\frac{1}{y-5}-\frac{1}{y+2}\right)dy=-\frac{3}{2}\frac{2x}{{x}^{2}+9}dx\\ ⇒\frac{1}{7}\int \left(\frac{1}{y-5}-\frac{1}{y+2}\right)dy=-\frac{3}{2}\int \frac{2x}{{x}^{2}+9}dx\\ ⇒\frac{1}{7}\left(\mathrm{ln}\left(y-5\right)-\mathrm{ln}\left(y+2\right)\right)=-\frac{3}{2}\mathrm{ln}\left({x}^{2}+9\right)+c\\ ⇒\mathrm{ln}\left(\frac{y-5}{y+2}{\right)}^{1/7}=\mathrm{ln}\left[\left({x}^{2}+9{\right)}^{-3/2}\cdot c\right]\\ ⇒\left(\frac{y-5}{y+2}{\right)}^{1/7}=\left({x}^{2}+9{\right)}^{-3/2}\cdot c\end{array}$