Linda Seales

2021-12-27

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

Daniel Cormack

Expert

Simplify the equation into the first order separable ODE
$\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)dy={y}^{2}dx$ (1)
$\left(1+{x}^{2}\right)\left(1+{y}^{2}\right)\frac{dy}{dx}={y}^{2}$
$\frac{\left(1+{y}^{2}\right)}{{y}^{2}}\frac{dy}{dx}=\frac{1}{\left(1+{x}^{2}\right)}$
$\left(\frac{1}{{y}^{2}}+1\right)\frac{dy}{dx}=\frac{1}{\left(1+{x}^{2}\right)}$
$\left(\frac{1}{{y}^{2}}+1\right)dy=\frac{1}{{x}^{2}+1}dx$
Integrate of both side
$\int \left(\frac{1}{{y}^{2}}+1\right)dy=\int \frac{1}{{x}^{2}+1}dx$
$\int \frac{1}{{y}^{2}}dy+\int 1\cdot dy=\int \frac{1}{{x}^{2}+1}dx$
$\frac{-1}{y}+y={\mathrm{tan}}^{-1}\left(x\right)+C$ (2)
Solve for y
$\frac{-1}{y}+y={\mathrm{tan}}^{-1}\left(x\right)+C$
$\frac{{y}^{2}-1}{y}={\mathrm{tan}}^{-1}\left(x\right)+C$
${y}^{2}-1=\left({\mathrm{tan}}^{-1}\left(x\right)+C\right)y$
${y}^{2}-\left({\mathrm{tan}}^{-1}\left(x\right)+C\right)y-1=0$
${y}^{2}-\left({\mathrm{tan}}^{-1}\left(x\right)+C\right)y-1=0$ (3)
The standard form of the solution of a quadratic equation $\left(a{x}^{2}+bx+c\right)$ is given as

poleglit3

Expert

Simplifying
$\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)\cdot dy={y}^{2}dx$
Reorder the terms:
$\left(1+{x}^{2}+{x}^{2}{y}^{2}+{y}^{2}\right)\cdot dy={y}^{2}dx$
Reorder the terms for easier multiplication:
$dy\left(1+{x}^{2}+{x}^{2}{y}^{2}+{y}^{2}\right)={y}^{2}dx$
$\left(1\cdot dy+{x}^{2}\cdot dy+{x}^{2}{y}^{2}\cdot dy+{y}^{2}\cdot dy\right)={y}^{2}dx$
Reorder the terms:
$\left({dx}^{2}y+{dx}^{2}{y}^{3}+1dy+{dy}^{3}\right)={y}^{2}dx$
$\left({dx}^{2}y+{dx}^{2}{y}^{3}+1dy+{dy}^{3}\right)={y}^{2}dx$
Solving
${dx}^{2}y+{dx}^{2}{y}^{3}+1dy+{dy}^{3}=dx{y}^{2}$
Solving for variable d.
Move all terms containing d to the left, all other terms to the right.
Add $-1left.dxright.{y}^{2}$ to each side of the equation.
${dx}^{2}y+{dx}^{2}{y}^{3}+1dy+-1dx{y}^{2}+{dy}^{3}=dx{y}^{2}+-1dx{y}^{2}$
Reorder the terms:

Vasquez

Expert

$\begin{array}{}\left(1+{x}^{2}+{y}^{2}+{x}^{2}{y}^{2}\right)dy={y}^{2}dx\\ \left(1+{x}^{2}\right)\left(1+{y}^{2}\right)dy={y}^{2}dx\\ \left(1+{y}^{2}\right)/{y}^{2}dy=1/\left(1+{x}^{2}\right)dx\\ \int \left(1+{y}^{2}\right)/{y}^{2}dy=\int 1/\left(1+{x}^{2}\right)dx\\ \int 1/{y}^{2}+1dy=\int 1/\left(1+{x}^{2}\right)dx\\ -1/y+y=\mathrm{arctan}\left(x\right)+C\end{array}$