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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS3 Marks
Question
Find the particular solution of the differential equation $(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}=0,\ \text{given that y}=1\ \text{when x}=0. $
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Answer

The given differential equation is
$(1+\text{e}^{2\text{x}})\text{dy}+(1+\text{y}^2)\text{e}^\text{x}\ \text{dx}=0$ $\text{or}\ \ (1+\text{e}^{2\text{x}})\text{dy}=-(1+\text{y}^2)\text{e}^{\text{x}}\ \text{dx}$
$\therefore\ \ \frac{1}{1+\text{y}^2}\text{dy}=-\frac{\text{e}^{\text{x}}}{1+\text{e}^{2\text{x}}}\text{dx}$
$\Rightarrow\ \ \int\frac{1}{1+\text{y}^2}\text{dy}=-\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}\text{dx}\ \ ...(1)$
$\text{Let l}=\int\frac{\text{e}^\text{x}}{1+\text{e}^{2\text{x}}}1+\text{dx}$
$\text{Put}\ \ \text{e}^\text{x}=\text{t},\ \therefore\ \text{e}^\text{x}\ \text{dx}=\text{dt}$
$\therefore\ \text{l}=\int\frac{\text{dt}}{1+\text{t}^2}=\tan^{-1}\text{t}=\tan^{-1}(\text{e}^\text{x})$
$\therefore\ \text{form}\ (2),\ \tan^{-1}\text{y}=\tan(\text{e}^\text{x})+\text{c}\ \ ...(2)$
$\text{Now}, \ \text{x}=0,\ \text{y}=1$
$\therefore\ \tan^{-1}1=-\tan(\text{e}^0)+\text{c}$ $\Rightarrow\ \ \frac{\pi}{4}=-\tan^{-1}(1)+\text{c}$
$\Rightarrow\ \ \frac{\pi}{4}=-\frac{\pi}{4}+\text{c}\ \ \Rightarrow\ \ \text{c}=\frac{\pi}{2}$
$\therefore\ \ \text{form}\ (2), \tan^{-1}=-\tan\text{e}^{\text{x}}+\frac{\pi}{2},$ which is required solution.

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