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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the differential equation:
$(\tan^{-1}\text{y} - x) \text{dy} = ( 1 + \text{y}^{2}) \text{dx}$
✓

Answer

Given differential equation can be writen as
$\frac{\text{dx}}{\text{dy}} + \frac{1}{1 + \text{y}^{2}} . \text{x} = \frac{\tan^{-1}\text{y}}{1 + \text{y}^{2}}$
$\therefore$ Intergrating factor is $e^{\tan^{-1}} \text{y}$
$\therefore$ Solution is: $\text{x}. e^{\tan^{-1}}\text{y} = \int \frac{\tan^{-1}{\text{y}}.e^{\tan^{-1}}\text{y}}{1 + \text{y}^{2}}\text{dy}$
$\Rightarrow \text{x .e}^{\tan^{-1}\text{y}} = \int \text{t}\text{ e}^{\text{t}}\text{ dt where } \tan^{-1}\text{y} = \text{t}$
$= \text{t e' - e' + c = e}^{\tan^{-1}\text{y}} (\tan^{-1}\text{y} - 1) + \text{c}$
$\text{or x} = \tan^{-1}\text{y} - 1 + \text{c e}^{-\tan^{-1}}\text{y}$

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