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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
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Answer

The given differential equation is $\big(\cot^{-1}\text{y} + \text{x}\big)\text{dy}= \big(1 + \text{y}^2\big) \text{dx}$
This differential equation can be written as
$\frac{\text{dx}}{\text{dy}} = \frac{\cot^{-1}\text{y}+\text{x}}{1+\text{y}^2}$
$\Rightarrow\frac{\text{dx}}{\text{dy}} + \Big(-\frac{1}{1+\text{y}^2}\Big)\text{x}= \frac{\cot^{-1}\text{y}}{1+\text{y}^2}$
This is a linear differential equation with $\text{P}=-\frac{1}{1+\text{y}^2}$ and $\text{Q}= \frac{\cot^{-1}\text{y}}{1+\text{y}^2}$
$\text{I}.\text{F}.=\text{e}^{-\int\frac{1}{1+\text{y}2}\text{dy}}=\text{e}^{\cot{^{-1}{\text{y}}}}$
Multiply the differential equation by integration factor (I.F.), we get
$\frac{\text{dx}}{\text{dy}}\text{e}^{\cot{^{-1}\text{y}}}-\frac{\text{x}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}=\frac{\cot^{-1}\text{y}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}$
$\Rightarrow\frac{\text{d}}{\text{dy}}\Big(\text{xe}^{\cot{^{-1}\text{y}}}\Big)=\frac{\cot^{-1}\text{y}}{\big(1+\text{y}^2\big)}\text{e}^{\cot{^{-1}\text{y}}}$
Integrating both sides with respect y, we get
$\text{xe}^{\cot{^{-1}\text{y}}}=\int\frac{\cot^{-1}\text{y}}{\big(1+{\text{y}^2\big)}}\text{e}^{\cot{^{-1}\text{y}}} \text{dy}+\text{C}$
Putting $\text{t}=\cot^{-1}\text{y}$ and $\text{dt}=-\frac{1}{1+\text{y}^2}\text{dy},$ we get
$\text{xe}^{\cot{^{-1}\text{y}}}=-\int \text{te}^{\text{t}}\text{dt}+\text{C}$
$\Rightarrow\text{xe}^{\cot{^{-1}\text{y}}}=-\text{e}\big(\text{t}-1\big)+\text{C}$
$\Rightarrow\text{xe}^{\cot{^{-1}\text{y}}}=\text{e}^{\cot{^{-1}\text{y}}}\big(1-\cot^{-1}\text{y}\big) + \text{C}$

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