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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
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Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}+\frac{1}{(\text{x}^2+1)^2}=0$
$\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}=-\frac{1}{(\text{x}^2+1)^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{4\text{x}}{\text{x}^2+1}$
$\text{Q}=-\frac{1}{(\text{x}^2+1)^2}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{2\int\frac{2\text{x}}{\text{x}^2+1}\text{dx}}$
$=\text{e}^{2\log|\text{x}^2+1|}$
$=(\text{x}^2+1)^2$
Multiplying both sides of (1) by $(x^2 + 1)^2$, we get
$(\text{x}^2+1)^2\Big(\frac{\text{dy}}{\text{dx}}+\frac{4\text{x}}{\text{x}^2+1}\text{y}\Big)=(\text{x}^2+1)^2\Big[-\frac{1}{(\text{x}^2+1)^2}\Big]$
$\Rightarrow\ (\text{x}^2+1)^2\frac{\text{dy}}{\text{dx}}+4\text{x}(\text{x}^2+1)\text{y}=-1$
Integrating both sides with respect to x, we get
$(\text{x}^2+1)^2\text{y}=-\int\text{dx + C}$
$\Rightarrow\ (\text{x}^2+1)^2\text{y}=-\text{x + C}$
Hence, $(\text{x}^2+1)^2\text{y}=-\text{x + C}$ is the required solution.

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