Question

$\frac{dy}{dx}+y\cos x=y^n\sin 2x.$

• A. $\frac{1}{y^{n+1}}=2\sin x-\frac{2}{1-n}+c{e}^{(n-1)\sin x}$
• B. $\frac{1}{y^{n-1}}=2\sin x+\frac{2}{1-n}+c{e}^{(n-1)\sin x}$
• C. $\frac{1}{y^{n-1}}=2\sin x-\frac{2}{1-n}+c{e}^{(n-1)\sin x}$
• D. $\frac{-1}{y^{n-1}}=2\sin x-\frac{2}{1-n}+c{e}^{(n-1)\sin x}$

Answer 1

Answer: (C)

$\frac{1}{y^{n-1}}=2\sin x-\frac{2}{1-n}+c{e}^{(n-1)\sin x}$

$\frac{dy}{dx}+y\cos x=y^n\sin 2x\Rightarrow \frac{1}{y^n}\frac{dy}{dx}+\frac{1}{y^{n-1}}\cos x=2\sin x\cos x$

Put $\frac{1}{y^{n-1}}=v\Rightarrow (1-n)\frac{1}{y^n}\frac{dy}{dx}=\frac{dv}{dx}$

$\therefore \frac{1}{1-n}\frac{dv}{dx}+v\cos x=2\sin x\cos x\Rightarrow \frac{dv}{dx}+v(1-n)\cos x=2(1-n)\sin x\cos x$ ...(1)

Here $P=(1-n)\cos x\Rightarrow \int PdP=\int (1-n)\cos xdx=(1-n)\sin x$

$\therefore I.F.=e^{(1-n)\sin x}$

Multiplying (1) by $I.F.$ we get

$e^{(1-n)\sin x}\frac{dv}{dx}+e^{(1-n)\sin x}.v(1-n)\cos x=e^{(1-n)\sin x}.2(1-n)\sin x\cos x$

Integrating both side, we get

$v{e}^{(1-n)\sin x}=\int 2(1-n)\sin x\cos x.e^{(1-n)\sin x}dx$

Put $(1-n)\sin x=t\Rightarrow (1-n)\cos xdx=dt$

$\therefore v{e}^t=\int 2\frac{t}{1-n}{e}^{t}dt=\frac{2}{1-n}{e}^t(t-1)+c\Rightarrow v=\frac{2}{1-n}(t-1)+c{e}^{-t}$

$\Rightarrow \frac{1}{y^{n-1}}=2\sin x-\frac{2}{1-n}+c{e}^{(n-1)\sin x}$