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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equations given in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin 2\text{x};\ \text{y}=2\ \text{when x}=\frac{\pi}{2}$
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Answer

The given differential equation is $\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin\ 2\text{x}.$This is a linear differential equation of the form:
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p}=-3\cot\text{x}\ \text{and}\ \text{Q}=\sin2\text{x})$
$\text{Now, I.F}=\text{e}^{\int\text{pdx}}=\text{e}^{-3\int\cot\text{x}\ \text{dx}}=\text{e}^{-3\log|\sin\text{x}|}=\text{e}^{\log|\frac{1}{\sin^3\text{x}}|}=\frac{1}{\sin^3\text{x}}.$
The general solution of the given differential equation is given by the relation,
$\text{y(I.F)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$
$\Rightarrow​​\text{y}\cdot\frac{1}{\sin^3\text{x}}=\int\Big[\sin2\text{x}.\frac{1}{\sin^3\text{x}}\text{dx}+\text{C}\Big]$
$\Rightarrow\text{y cosec}^3\ \text{x}=2\int(\cot\text{x}\ \text{cosec}\ \text{x})\text{dx}+\text{C}$
$\Rightarrow\text{y cosec}^3\ \text{x}=-2\text{cosec}\ \text{x}+\text{C}$
$\Rightarrow\text{y}=-\frac{2}{\text{cosec}^2\text{x}}+\frac{\text{C}}{\text{cosec}^3\text{x}}$
$\Rightarrow\text{y}=-2\sin^2\text{x}+\text{C}\sin^3\text{x}\ \ ...(1)$
$\text{Now,}\ \text{y}=2\ \text{at}\ \text{x}=\frac{\pi}{2}.$
Therefore, we get:
$2=-2+\text{C}$
$\Rightarrow\text{C}=4$
Substituting C = 4 in equation (1), we get:
$\text{y}=-2\sin^2\text{x}+4\sin^3\text{x}$
$\Rightarrow\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$
This is the required particular solution of the given differential equation.

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