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DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the differential equation $\text{dy}=\cos\text{x}(2-\text{y}\text{cosesx})\text{dx}$ given that $\text{y}=2$ when $\text{x}=\frac{\pi}{2}.$
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Answer

We have
$\text{dy}=\cos\text{x}(2-\text{y}\text{cosesx})\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\cos\text{x}-\text{y}\text{cosec}\text{x}.\cos\text{x}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x}$
This is a linear differential equation.
On comparinvg it with $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q},$ we get
$\text{P}=\cot\text{x},\text{Q}=2\cos\text{x}$
$\text{I.F.}=\text{e}^{\int\text{Pdx}}=\text{e}^{\int\cot\text{xdx}}$
$=\text{e}^{\log\sin\text{x}}=\sin\text{x}$
Thus, the general solution is,
$\text{y}.\sin\text{x}=\int2.\cos\text{x}.\sin\text{x}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\sin\text{x}=\int\sin2\text{x}\text{dx}+\text{C}$
$\Rightarrow\text{y}.\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}$
Given that when $\text{x}=\frac{\pi}{2}$ and y = 2
$\Rightarrow2.\sin\frac{\pi}{2}=-\frac{\cos\pi}{2}+\text{C}$
$\Rightarrow2=\frac{1}{2}+\text{C}$
$\Rightarrow\text{C}=\frac{3}{2}$
On substituting the value of C in Eq. (i), we get
$\text{y}\sin\text{x}=-\frac{1}{2}\cos2\text{x}+\frac{3}{2}$

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