Solve the following differential equation: ( ) dy dx +y =2 ^2x - Vidyadip

Question

Solve the following differential equation:
$(\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$

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Answer

We have,
$(\sin\text{x})\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
$\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}=2\sin\text{x}\cos\text{x}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\cot\text{x}$
$\text{Q}=2\sin\text{x}\cos\text{x}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\cot\text{xdx}}$
$=\text{e}^{\log|\sin\text{x}|}=\sin\text{x}$
Multiplying both sides of (1) by $\sin\text{x},$ we get
$\sin\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\cot\text{x}\Big)=\sin\text{x}\times2\sin\text{x}\cos\text{x}$
$\sin\text{x}\frac{\text{dy}}{\text{dx}}+\text{y}\cos\text{x}=2\sin^2\text{x}\cos\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sin\text{x}=2\int\sin^2\text{x}\cos\text{x dx + C}\ \dots(2)$
Putting $\sin\text{x}=\text{t}$
$\Rightarrow\ \cos\text{x dx = dt}$
Therefore, (2) becomes
$\text{y}\sin\text{x}=2\int\text{t}^2\text{dt + C}$
$\Rightarrow \text{y}\sin\text{x}=\frac{2}3\text{t}^3+\text{C}$
$\Rightarrow \text{y}\sin\text{x}=\frac{2}3\sin^3\text{x}+\text{C}$
Hence, $\text{y}\sin\text{x}=\frac{2}3\sin^3\text{x}+\text{C}$ is the required solution.

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