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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the particular solution of the differential equation $(1 + \text{x}^{2}) \frac{\text{dy}}{\text{dx}} = \text{(e}^{\text{m}\tan^{-1}\text{x}} -\text{y}),$ given that $\text{y = 1 when x = 0}$
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Answer

Given differential equation can be written as
$\frac{\text{dy}}{\text{dx}} + \frac{1}{1 + \text{x}^{2}}\text{y} = \frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1 + \text{x}^{2}}$
Intergrating factor is ${\text{e}^{\text{m}\tan{-1}\text{x}}} $
Solution is $\text{y}.{\text{e}^{\tan^{-1}\text{x}}} = \int\frac{\text{e}^{\text{m}\tan^{-1}\text{x}}}{1 + \text{x}^{2}} .{\text{e}^{\tan^{-1}\text{x}}} \text{dx}$
$\Rightarrow\text{y e}^{\tan^{-1}}\text{x} = \int\text{e}^{(\text{m + 1 )t}}\text{dt}, \text{where} \tan^{-1}\text{x = t} $
$= \frac{\text{e}^{(\text{m + 1) t}}}{\text{m} + 1} = \frac{\text{e}^{(\text{m} + 1)\tan^{-1} \text{x}}}{\text{m + 1}} + \text{c}$
$\text{y = 1, x = 0}\Rightarrow\text{c} = \frac{\text{m}}{\text{m + 1}}$
$\text{y e}^{\tan^{-1}\text{x}} = \frac{\text{e}^{(\text{m} + 1)\tan^{-1} \text{x}}}{\text{m + 1}} + \frac{\text{m}}{\text{m + 1}}$

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