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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

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

Answer

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

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