Question

The solution of $\frac{dy}{dx}=\frac{e^x({\sin }^{2}x+\sin 2x)}{y(2\log y+1)}$ is-

• A. $y^2\left(\log y\right)-e^x{\sin }^{2}x+c=0$
• B. $y^2\left(\log y\right)-e^x{\cos }^{2}x+c=0$
• C. $y^2\left(\log y\right)+e^x{\cos }^{2}x+c=0$
• D. None of these

Answer 1

The correct option is A $y^2\left(\log y\right)-e^x{\sin }^{2}x+c=0$

Rearranging above one, we get,

$y(2\log y+1)dy=e^x({\sin }^{2}x+\sin 2x)dx$

Integrating both sides,

$\int y(2\log y+1)dy=\int e^x({\sin }^{2}x+\sin 2x)dx$

$\int 2y\log ydy+\int ydy=\int e^x{\sin }^{2}xdx+\int e^x\sin 2xdx$

Solving $L.H.S$ using integration by parts for first part,

$\log y\int 2ydy-\int (\frac{d\log y}{dy}\int (2y\left)dy\right)dy+\int ydy$

$\log y\times \frac{2y^2}{2}-\int \frac{1}{y}\frac{2y^2}{2}dy+\int ydy$

$y^2\log y-\int ydy+\int ydy=y^2\log y$

Solving $R.H.S$ using integration by parts for first part,

${\sin }^{2}x\int e^{x}dx-\int (\frac{d{\sin }^{2}x}{dx}\int e^{x}dx)dx+\int e^x\sin 2xdx$

${\sin }^{2}x{e}^x-\int 2\sin x\cos x{e}^{x}dx+\int e^x\sin 2xdx$

As $2\sin x\cos x=\sin 2x$, replace it in above,

$={\sin }^{2}x{e}^x-\int \sin 2x{e}^{x}dx+\int e^x\sin 2xdx=e^x{\sin }^{2}x$

$\therefore$ solution to the differential equation will be,

$y^2\log y=e^x{\sin }^{2}x+C$