# Question #78e39

Jan 8, 2017

$\text{The General Solution : } y + 2 {\sin}^{2} x = c {\sin}^{3} x .$

$\text{The Particular Solution : } y + 2 {\sin}^{2} x = 4 {\sin}^{3} x .$

#### Explanation:

This is a Linear Differential Equation of First Order , usually

taken as, $\frac{\mathrm{dy}}{\mathrm{dx}} + y P \left(x\right) = Q \left(x\right)$.

To find its General Soln. , we need to multiply it by the Integrating

Factor, i.e., IF , given by, IF $= {e}^{\int P \left(x\right) \mathrm{dx}}$.

In our Example, $P \left(x\right) = - 3 \cot x , s o , \int P \left(x\right) \mathrm{dx} = \int \left\{- 3 \cot x\right\} \mathrm{dx}$

$= - 3 \ln \sin x = \ln {\sin}^{- 3} x \Rightarrow {e}^{P \left(x\right) \mathrm{dx}} = {e}^{\ln {\sin}^{- 3} x} = \frac{1}{\sin} ^ 3 x$.

Multiplying the Diff. Eqn., by $\frac{1}{\sin} ^ 3 x$, we get,

${\sin}^{- 3} x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y \cot \frac{x}{\sin} ^ 3 x = \frac{\sin 2 x}{\sin} ^ 3 x$

$\therefore {\sin}^{- 3} x \frac{\mathrm{dy}}{\mathrm{dx}} - 3 y \cos x {\sin}^{- 4} x = \frac{2 \sin x \cos x}{\sin} ^ 3 x = \frac{2 \cos x}{\sin} ^ 2 x$

$\therefore {\sin}^{- 3} x \frac{d}{\mathrm{dx}} \left(y\right) + y \frac{d}{\mathrm{dx}} \left({\sin}^{- 3} x\right) = 2 \cot x \csc x$

$\therefore \frac{d}{\mathrm{dx}} \left\{y {\sin}^{- 3} x\right\} = 2 \cot x \csc x$

Integrating, $y {\sin}^{- 3} x = 2 \int \cot x \csc x + c = - 2 \csc x + c$

$\therefore y + 2 {\sin}^{2} x = c {\sin}^{3} x ,$ is the reqd. Gen. Soln.

To find the Particular Soln. , we utilise the Initial Condition :

$y = 2 , \text{ when } x = \frac{\pi}{2}$.

Sub.ing in the Gen. Soln., we get, $2 + 2 = c = 4$

Hene, the P.S. $: y + 2 {\sin}^{2} x = 4 {\sin}^{3} x$

Enjoy Maths.!