Question

\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

Answer 1

We have,
\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]
\[ \Rightarrow dy = \left( \cos^3 x \sin^2 x + x\sqrt{2x + 1} \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( \cos^3 x \sin^2 x + x\sqrt{2x + 1} \right)dx\]
\[ \Rightarrow y = \int \cos^3 x \sin^2 x dx + \int x\sqrt{2x + 1}dx \]
\[ \Rightarrow y = I_1 + I_2 . . . . . \left( 1 \right)\]
where 
\[ I_1 = \int \cos^3 x \sin^2 x dx \]
\[ I_2 = \int x\sqrt{2x + 1}dx\]
Now,
\[ I_1 = \int \cos^3 x \sin^2 x dx\]
\[ = \int \sin^2 x \left( 1 - \sin^2 x \right)\cos x dx\]
\[\text{Putting }t = \sin x,\text{ we get }\]
\[dt = \cos x dx\] 
\[ \Rightarrow I_1 = \int t^2 \left( 1 - t^2 \right)dt\]
\[ = \int\left( t^2 - t^4 \right)dt\]
\[ = \frac{t^3}{3} - \frac{t^5}{5} + C_1 \]
\[ = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C_1 \]
\[ I_2 = \int x\sqrt{2x + 1}dx\]
\[\text{Putting }t^2 = 2x + 1, \text{ we get }\]
\[2t dt = 2dx\]
\[ \Rightarrow tdt = dx\]
Now,
\[ I_2 = \int\left( \frac{t^2 - 1}{2} \right)t \times t dt\]
\[ = \frac{1}{2}\int\left( t^4 - t^2 \right) dt\]
\[ = \frac{t^5}{10} - \frac{t^3}{6} + C_2 \]
\[ = \frac{\left( 2x + 1 \right)^\frac{5}{2}}{10} - \frac{\left( 2x + 1 \right)^\frac{3}{2}}{6} + C_2\]
\[\text{Putting the values of }I_1\text{ and }I_2 \text{ in }\left( 1 \right), \text{ we get }\]
\[y = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + C_1 + \frac{\left( 2x + 1 \right)^\frac{5}{2}}{10} - \frac{\left( 2x + 1 \right)^\frac{3}{2}}{6} + C_2 \]
\[y = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + \frac{\left( 2x + 1 \right)^\frac{5}{2}}{10} - \frac{\left( 2x + 1 \right)^\frac{3}{2}}{6} + C ...............\left( \text{Where, } C = C_1 + C_2 \right)\]
\[\text{ Hence, }y = \frac{\sin^3 x}{3} - \frac{\sin^5 x}{5} + \frac{\left( 2x + 1 \right)^\frac{5}{2}}{10} - \frac{\left( 2x + 1 \right)^\frac{3}{2}}{6} +\text{ C is the solution to the given differential equation.}\]