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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Find the particular solution of the differential equation $\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x},$ given that $\text{y}=0\ \text{when x}=\frac{\pi}{3}.$
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Answer

$\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x}$
Differential equation is of the form
$\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}$
$\frac{\text{dy}}{\text{dx}}+2\text{y}\ \tan\text{x}=\sin\text{x}\ \ ....(1 )$
$\text{Where P}=2\tan\text{x}\ \&\ \text{Q}=\sin\text{x}$
$\text{IF}=\text{e}^{\int\text{p dx}}$
$\text{IF}=\text{e}^{\int2\tan\text{x dx}}$
$\text{IF}=\text{e}^{2\log\sec\text{x}}$
$\text{IF}=\text{e}^{\log\sec^2\text{x}}$
$\text{IF}=\sec^2\text{x}$
$\text{y}(\text{IF})=\int(\text{Q}\times\text{IF})\text{dx}+\text{c}$
$\text{y}(\sec^2\text{x})=\int\sin\text{x}\sec^2\text{x dx}+\text{c}$
$\text{y}\sec^2\text{x}=\int\sin\text{x}\frac{1}{\cos^2\text{x}}\text{dx}+\text{C}$
$\text{y}\sec^2\text{x}=\int\frac{\sin\text{x}}{\cos\text{x}}\times\frac{1}{\cos\text{x}}\text{dx}+\text{C}$
$\text{y}\sec^2\text{x}=\int\tan\text{x}\sec\text{x dx}+\text{C}$
$\text{y}\sec^2\text{x}=\sec\text{x}+\text{C}$
$\text{y}=\frac{\sec\text{x}}{\sec^2\text{x}}+\frac{\text{c}}{\sec^2\text{x}}$
$\text{y}=\cos\text{x}+\text{C}\cos^2\text{x}\ \ ....(2)$
$\text{Putting x}=\frac{\pi}{3}\ \&\ \text{y}=0$
$0=\cos\frac{\pi}{3}+\text{C}\cos^2\frac{\pi}{3}$
$0=\frac{1}{2}+\text{C}\Big(\frac{1}{4}\Big)^2$
$\frac{-1}{2}=\text{C}\Big(\frac{1}{4}\Big)$
$\frac{-4}{2}=\text{C}$
$\text{C}=-2$
Putting value of C in (1)
$\text{y}=\cos\text{x}+\text{C}\cos^2\text{x}$
$\text{y}=\cos\text{x}-2\cos^2\text{x}$

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