Question

Solve the given differential equation.

$\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{8}\mathbf{xy}\mathbf{)}\mathbf{dx}$

Short answer

The general solution of the given differential equation is $y=\frac{1}{4}-\frac{c_2}{{(x^2+4)}^4}$

Step-by-step solution

Use the concept of variable separable form

$\left({\mathrm{x}}^2+4\right)\mathrm{dy}=2\mathrm{x}(1-4\mathrm{y})\mathrm{dx}\mathrm{Factor}\mathrm{out}2\mathrm{x}$

Rewrite the equation and solve using variable separable form

$\begin{array}{rcl}\left({\mathrm{x}}^2+4\right)\mathrm{dy}& =& 2\mathrm{x}(1-4\mathrm{y})\mathrm{dx}\mathrm{Factorout}2\mathrm{x}\\ \frac{\mathrm{dy}}{1-4\mathrm{y}}& =& \frac{2\mathrm{x}}{{\mathrm{x}}^2+4}\mathrm{dx}\end{array}$

Integrating both sides

$\int \frac{dy}{1-4y}=\int \frac{2x}{x^2+4}dx$

$-\frac{1}{4}\ln |1-4y|+\ln c=\ln |x^2+4|$

$-\ln |1-4y|+4\ln c=4\ln |x^2+4|$

$\ln |1-4y|+4\ln |x^2+4|=4\ln \left|c\right|$

Applying properties

$ln|1-4y|+ln\left|{(x^2+4)}^4\right|=ln\left|c^4\right|alnb=ln{b}^a$

$\ln \left|(1-4y){(x^2+4)}^4\right|=\ln \left|c^4\right|\ln a+\ln b=\ln ab$

$(1-4y){(x^2+4)}^4=c^4$

$1-4y=\frac{c_1}{{(x^2+4)}^4}{c}_1=c^4$

$4y=1-\frac{c_1}{{(x^2+4)}^4}$

$:\#\mathbf{b}1\mathbf{badf};>\mathbf{y}=\frac{1}{4}-\frac{c_2}{{(x^2+4)}^4}{c}_2=\frac{c_1}{4}$