Question

The solution of the differential equation $e^{-x}dy(y+1)dy+(Co{s}^{2}x-Sin2x)y(dx)=0$ subjected to the condition that $y=1$ when $x=0$ is

• A. $(y+1)+e^{x}Co{s}^{2}x=2$
• B. $y+Logy-e^{x}Co{s}^{2}x=1$
• C. $Log(y+1)+e^{x}Co{s}^{2}x=1$
• D. $y+Logy+e^{x}Co{s}^{2}x=2$

Answer 1

Answer: (D)

We have
$e^{-x}(y+1)dy+\left({\cos }^{2}x-\sin 2x\right)ydx=0$
$\Rightarrow \frac{(y+1)}{y}dy=-e^x\left({\cos }^{2}x-\sin 2x\right)dx$
$\Rightarrow \left(1+\frac{1}{y}\right)dy=-e^x\left({\cos }^{2}x-\sin 2x\right)dx$
On integrating both sides, we get
$y+\log y=-e^x{\cos }^{2}x+\int e^x\sin 2xdx$
$-\int e^x\sin 2xdx+c$
$\Rightarrow y+\log y=-e^x{\cos }^{2}x+c$
At $x=0,y=1$
$\Rightarrow 1+0=-e^0\cos 0+c$
$\Rightarrow c=2$
$\therefore$ Required solution is
$y+\log y=-e^x{\cos }^{2}x+2$