aspifsGak5u

2021-12-26

Solve the DE: $\left({x}^{2}+1\right)dx+{x}^{2}{y}^{2}dy=0$

chumants6g

Step 1
Given differential equation is $\left({x}^{2}+1\right)dx+{x}^{2}{y}^{2}dy=0$
Here we solve above differential equation
Step 2
Now $\left({x}^{2}+y\right)dx+{x}^{2}{y}^{2}dy=0$
$⇒\left(\frac{{x}^{2}+1}{{x}^{2}}\right)dx+{y}^{2}dy=0$
$⇒\left(1+\frac{1}{{x}^{2}}dx+{y}^{2}dy=0$
$⇒\left(1+{x}^{-2}\right)dx+{y}^{2}dy=0$
Integrating, $\int \left(1+{x}^{-2}\right)dx+\int {y}^{2}dy=c$
$⇒x+\frac{{x}^{-2+1}}{-2+1}+\frac{{y}^{2+1}}{2+1}=c$
$⇒x+\frac{{x}^{-1}}{-1}+\frac{{y}^{3}}{3}=c$
$⇒x-\frac{1}{x}+\frac{{y}^{3}}{3}=c$ whose c is arbitrary constant.

Tiefdruckot

$y\left(x\right)=\frac{\left(-\sqrt[3]{3}-{3}^{\frac{5}{6}}i\right)\sqrt[3]{{C}_{1}-x+\frac{1}{x}}}{2}$
$y\left(x\right)=\frac{\left(-\sqrt[3]{3}+{3}^{\frac{5}{6}}i\right)\sqrt[3]{{C}_{1}-x+\frac{1}{x}}}{2}$
$y\left(x\right)=\sqrt[3]{{C}_{1}-3x+\frac{3}{x}}$

user_27qwe

$\begin{array}{}\left({x}^{2}+1\right)dx+{x}^{2}{y}^{2}dy=0\\ \left({x}^{2}+y\right)dx+{x}^{2}{y}^{2}dy=0\\ ⇒\left(\frac{{x}^{2}+1}{{x}^{2}}\right)dx+{y}^{2}dy=0\\ ⇒\left(1+\frac{1}{{x}^{2}}dx+{y}^{2}dy=0\\ ⇒\left(1+{x}^{-2}\right)dx+{y}^{2}dy=0\\ \int \left(1+{x}^{-2}\right)dx+\int {y}^{2}dy=c\\ ⇒x+\frac{{x}^{-2+1}}{-2+1}+\frac{{y}^{2+1}}{2+1}=c\\ ⇒x+\frac{{x}^{-1}}{-1}+\frac{{y}^{3}}{3}=c\\ ⇒x-\frac{1}{x}+\frac{{y}^{3}}{3}=c\end{array}$