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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
For each of the differential equations given in find a particular solution satisfying the given condition:
$\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x};\text{y}=0\ \text{when x}=\frac{\pi}{3}$
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Answer

The given differential equation is $\frac{\text{dy}}{\text{dx}}+2\text{y}\tan\text{x}=\sin\text{x}.$ This is a linear equation of the form: $\frac{\text{dy}}{\text{dx}}+\text{py}=\text{Q}\ (\text{where p}=2\tan\text{x}\ \text{and} \ \text{Q}=\sin\text{x})$ $\text{Now, I.F}=\text{e}^{\int\text{pdy}}=\text{e}^{\int2\tan\text{x}\ \text{dx}}=\text{e}^{2\log|\sec\text{x}|}=\text{e}^{\log(\sec^2\text{x})}=\sec^2\text{x}.$ The general solution of the given differential equation is given by the relation, $\text{y(I.F.)}=\int(\text{Q}\times\text{I.F.})\text{dx}+\text{C}$ $\Rightarrow​​\text{y}(\sec^2\text{x})=\int(\sin\text{x}\cdot\sec^2\text{x})\text{dx}+\text{C}$ $\Rightarrow\text{y}\sec^2\text{x}=\int(\sec\text{x}\cdot\tan\text{x})\text{dx}+\text{C}$ $\Rightarrow\text{y}\sec^2\text{x}=\sec\text{x}+\text{C}\ \ ....(1)$ $\text{Now, y}=0\ \text{at x} =\frac{\pi}{3}\cdot$ Therefore, $0\times\sec^2\frac{\pi}{3}=\sec\frac{\pi}{3}+\text{C}$ $\Rightarrow0=2+\text{C}$ $\Rightarrow\text{C}=-2$Substituting C = -2 in equation (1), we get:
$​​\text{y}\sec^2​​\text{x}=\sec​​\text{x}-2$ $\Rightarrow\text{y}=\cos\text{x}-2\cos^2\text{x}$Hence, the required solution of the given differential equation is $\text{y}=\cos\text{x}-2\cos^2\text{x}.$

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