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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following 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

We have, $(1+\text{y}^2)+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=0$ $\Rightarrow\ (\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$ $\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\frac{(1+\text{y}^2)}{(\text{x}-\text{e}^{\tan^{-1}\text{y}})}$$\Rightarrow\ \frac{\text{dx}}{\text{dy}}=-\frac{\text{x}-\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\ \frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ where $\text{P}=\frac{1}{1+\text{y}^2}$ $\text{Q}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\therefore$ I.F. $=\text{e}^{\int\text{Pdy}}$ $=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$ $=\text{e}^{{\tan^{-1}\text{y}}}$ Multiplying both sides of (1) by $​​​​\text{e}^{\tan^{-1}\text{y}},$ we get $\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ $\Rightarrow\ \text{e}^{\tan^{-1}\text{y}}\frac{\text{dx}}{\text{dy}}+\frac{\text{x}\text{e}^{\tan^{-1}\text{x}}}{1+\text{y}^2}=\frac{\text{e}^{\tan^{-1}\text{y}}}{1+\text{y}^2}$ Integrating both sides with respect to y, we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy + C}$ $\Rightarrow\ \text{x}\text{e}^{\tan^{-1}\text{y}}=\text{I + C}\ \dots(2)$ Here, $\text{I}=\int\frac{\text{e}^{2\tan^{-1}\text{y}}}{1+\text{y}^2}\text{dy}$ Putting $\tan^{-1}\text{y = t},$ we get $\frac{1}{1+\text{y}^2}\text{dy = dt}$ $\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$ $=\frac{\text{e}^{2\text{t}}}{2}$ $=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}$ Putting the value of I in (2), we get $\text{x}\text{e}^{\tan^{-1}\text{y}}=\frac{\text{e}^{2\tan^{-1}\text{y}}}{2}+\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+2\text{C}$ $\Rightarrow\ 2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ (where K = 2C) Hence, $2\text{x}\text{e}^{\tan^{-1}\text{y}}=\text{e}^{2\tan^{-1}\text{y}}+\text{K}$ is the required solution.

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