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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Find the particular solution of the differential equation $\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y},\text{ y}\neq0$ given that x = 0 when $\text{y}=\frac{\pi}{2}$
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Answer

We have,
$\frac{\text{dx}}{\text{dy}} + \text{x}\cot \text{y}=2\text{y} + \text{y}^{2} \cot \text{y}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dx}}{\text{dy}}+\text{Px}=\text{Q}$
Where $\text{P}=\cot\text{y}$ and $\text{Q}=2\text{y} + \text{y}^{2} \cot \text{y}$
$\therefore\ \text{I}.\text{F}.=\text{e}^{\int\text{P}\text{dy}}$
$=\text{e}^{\int\cot\text{y}\text{ dy}}$
$=\text{e}^{\log|\sin\text{y}|}$
$=\sin\text{y}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{y},$ we get
$\sin\text{y}\Big(\frac{\text{dx}}{\text{dy}}+\text{x}\cot\text{y}\Big)=\sin\text{y}\big(\text{y}^2\cot\text{y} + 2\text{y}\big)$
$\Rightarrow\sin \text{y}\frac{\text{dx}}{\text{dy}}+\text{x}\cos\text{y}=\text{y}^2\cos\text{y}+2\text{y}\sin\text{y}$
Integrating both sides with respect to y, we get
$\text{x}\sin \text{y}=\int\text{y}^{2}\cos\text{y}\text{ dy}+\int2\text{y}\sin\text{y}\text{ dy}+\text{C}$
$\Rightarrow\text{x}\sin\text{y}=\text{y}^2\int\cos\text{y dy}-\int\Big[\frac{\text{d}}{\text{dy}}(\text{y}^2)\int\cos\text{y dy}\Big]\text{dy}+\int2\text{y}\sin\text{y dy}+\text{C}$
$ \Rightarrow\text{x} \sin\text{y}=\text{y}^2 \sin\text{y}-\int2\text{y}\sin\text{y} + \int2\text{y}\sin\text{y}\text{ dy} + \text{C}$
$\Rightarrow\text{x}\sin \ \text{y}=\text{y}^2 \sin \ \text{y} + \text{C}$
Now,
$\therefore 0\times\sin \frac{\pi}{2}=\frac{\pi^2}{4} \sin \frac{\pi}{2}+\text{C}$
$\Rightarrow\text{C}=-\frac{\pi^2}{4}$
Putting the value of C, we get
$\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$
Hence, $\text{x}\sin \ \text{y}=\text{y}^2\sin \text{y}-\frac{\pi^2}{4}$ is the required solution.

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