Solve the following initial value problems: dy= (2-y cosecx)dx - Vidyadip

Question

Solve the following initial value problems:
$\text{dy}=\cos\text{x}(2-\text{y cosecx})\text{dx}$

✓

Answer

We have,
$\text{dy}=\cos\text{x}(2-\text{y cosecx})\text{dx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=2\cos\text{x}-\text{y}\cot\text{x}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\cos\text{x}\ ...(1)$
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
$\text{y}\sin\text{x}=\int\sin2\text{x dx}+\text{C}$
$\Rightarrow\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}$
Hence, $\text{y}\sin\text{x}=-\frac{\cos2\text{x}}{2}+\text{C}$ is the required solution.

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