Solve the following differential equation: (1+x^2) dy dx -2xy=(x^… - Vidyadip

Question

Solve the following differential equation:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}-2\text{xy}=(\text{x}^2+2)(\text{x}^2+1)$

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Answer

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

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