Solve the following differential equation: x dy=(2y+2x^4+x^2)dx

Question

Solve the following differential equation:
$\text{x dy}=(2\text{y}+2\text{x}^4+\text{x}^2)\text{dx}$

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Answer

Here, $\text{x dy}=(2\text{y}+2\text{x}^4+\text{x}^2)\text{dx}$
$\text{x}\frac{\text{dy}}{\text{dx}}=2\text{y}+2\text{x}^4+\text{x}^2$
$\frac{\text{dy}}{\text{dx}}-\frac{2}{\text{x}}\text{y}=2\text{x}^3+\text{x}$
It is a linear differential equation. Comparing it with equation,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{2}{\text{x}},\text{Q}=2\text{x}^3+\text{x}$
I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-2\int\frac{1}{1+\text{x}}\text{dx}}$
$=\text{e}^{-2\log|\text{x}|}$
$=\text{e}^{\log\big(\frac{1}{\text{x}^2}\big)}$
$=\frac{1}{\text{x}^2}$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}^2}\Big)=\int\big(2\text{x}^3+\text{x}\big)\Big(\frac{1}{\text{x}^2}\Big)\text{dx + C}$
$\frac{\text{y}}{\text{x}^2}=\int\Big(2\text{x}+\frac{1}{\text{x}}\Big)\text{dx + C}$
$\frac{\text{y}}{\text{x}^2}=2\frac{\text{x}^2}2+\log|\text{x}|+\text{C}$
$\text{y}=\text{x}^4+\text{x}^2\log|\text{x}|+\text{Cx}^2$

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