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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the general solution of the differential equation $(1+\text{y}^2)+(\text{x}-\text{e}^{{\tan^{-1}\text{y}}})\frac{\text{dy}}{\text{dx}}=0.$
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Answer

Given, differential equation is
$(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$
$\Rightarrow(1+\text{y}^2)\frac{\text{dy}}{\text{dx}}+\text{x}-\text{e}^{\tan^{-1}\text{y}}=0$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
This is a linear differential equation.
On comparing it with $\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q},$ we get
$\text{P}=\frac{1}{1+\text{y}^2},\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\text{I.F.}=\text{e}^{\int\text{P}\text{dx}}$
$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dx}}$
$=\text{e}^{\tan^{-1}\text{y}}$
So, the general solution is,
$\text{x}.\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}.\text{e}^{\tan^{-1}\text{y}}\text{dx}+\text{C}$
Put $\text{e}^{\tan^{-1}\text{y}}=\text{t}$
$\Rightarrow\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}=\text{dt}$
$\therefore\text{x}.\text{e}^{\tan^{-1}\text{y}}=\int\text{t}\text{dt}+\text{C}$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{t}^2}{2}+\text{C}$

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