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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x},\text{ y}=2,\text{ when x}=\frac{\pi}{2}$
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Answer

We have,
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x}\ ....(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-3\cot\text{x}$ and $\text{Q}=\sin2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-3\int\cot\text{x dx}}$
$=\text{e}^{-3\log|\sin\text{x}|}$
$=\text{cosec}^3\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\text{cosec}^3\text{x},$ we get
$\text{cosec}^3\text{x}\Big(\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}\Big)=\sin2\text{x}(\text{cosec}^3\text{x})$
$\Rightarrow\text{cosec}^3\text{x}\Big(\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}\Big)=2\cot\text{x cosec x}$
Integrating both sides with respect to x, we get
$\text{y }\text{cosec}^3\text{x}=2\int\cot\text{x}\text{ cosec}\text{ x dx}+\text{C}$
$\Rightarrow\text{y }\text{cosec}^3\text{x}=-2\text{cosec}\text{ x}+\text{C}$
$\Rightarrow\text{y}=-2\sin^2\text{x}+\text{C}\sin^3\text{x}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=2$
$\therefore\ 2=-2\sin^2\frac{\pi}{2}+\text{C}\sin^3\frac{\pi}{2}$
$\Rightarrow\text{C}=4$
Putting the value of C in (2), we get
$\text{y}=-2\sin^2\text{x}+4\sin^3\text{x}$
$\Rightarrow\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$
Hence, $\text{y}=4\sin^3\text{x}-2\sin^2\text{x}$ is the required solution.

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