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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
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Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x},\text{ y}\Big(\frac{\pi}{2}\Big)=0$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=\cot\text{x}$ and $\text{Q}=2\cos\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{x dx}}$
$=\text{e}^{\log\sin\text{x}}$
$=\sin\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=2\sin\text{x }\cos\text{x}$
$\Rightarrow\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=\sin2\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=\int\sin2\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}\ ...(2)$
Now,
$\text{y}\Big(\frac{\pi}{2}\Big)=0$
$\therefore\ 0\times\sin\frac{\pi}{2}=-\frac{\cos\pi}{2}+\text{C}$
$\Rightarrow\text{C}=-\frac{1}{2}$
Putting the value of C in (2), we get
$\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}-\frac{1}{2}$
$\Rightarrow2\text{y}\sin\text{x}=-(1+\cos2\text{x})$
$\Rightarrow2\text{y}\sin\text{x}=-2\cos^2\text{x}$
$\Rightarrow\text{y}=-\cot\text{x}\cos\text{x}$
Hence, $\text{y}=-\cot\text{x}\cos\text{x}$ is the required solution.

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