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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:
${(\text{x}^{2}-1)}\frac{\text{dy}}{\text{dx}}+\text{2xy}=\frac{1}{\text{x}^{2}-1};|\text{x}|\neq1$.
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Answer

Given differential equation can be written as
$\frac{\text{dy}}{\text{dx}}+\frac{{\text{2x}}}{{\text{x}^{2}-1}}\cdot{\text{y}}=\frac{1}{(\text{x}^{2}-\text{1})^{2}}$
Which is of the form $\frac{\text{dy}}{\text{dx}}+\text{P(x)}\cdot\text{y = Q(x)}$

$\int\text{P(x) dx}=\int\frac{\text{2x}}{\text{x}^{2}-1}\text{dx}=\log|\text{x}^{2}-1|$
$\therefore$ Integrating factor = $\text{e}^{\int\text{p(x) dx}}=\text{(x}^{2}-1)$
$\therefore$ The solution is $(x^2 - 1).Y = \int\frac{1}{\text{(x}^{2}-1)^{2}}\text{(x}^{2}-1)\text{ dx}$

$\text{(x}^{2}-1)\cdot\text{y}=\frac{1}{2}\log\Bigg|\frac{\text{x - 1}}{\text{x + 1}}\Bigg|+\text{c}$.

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