Question

The solution of the differential equation $e^{-x}(y+1)dy+(co{s}^{2}x-sin2x)ydx=0$ subject to the conditions $y\left(0\right)=1$

• A. $\log (y+1)+e^x{\cos }^{2}x=1$
• B. $y+\log y+e^x{\cos }^{2}x=2$
• C. $(y+1)+e^x{\cos }^{2}x=2$
• D. $y+\log y=e^2{\cos }^{2}x$

Answer 1

The correct option is B $y+\log y+e^x{\cos }^{2}x=2$

$e^{-x}(y+1)dy+({\cos }^{2}x-\sin 2x)ydx=0,y\left(0\right)=1$

$\left(\frac{y+1}{y}\right)dy+e^x({\cos }^{2}x-\sin 2x)dx=0$

$\Rightarrow (1+\frac{1}{y})dy+(e^x{\cos }^{2}x-e^x\sin 2x)dx=0$

$\Rightarrow d(y+\log y)+d\left(e^x{\cos }^{2}x\right)=0$

$\Rightarrow d(y+\log y)+d\left(e^x{\cos }^{2}x\right)=0$

Integrating the above equation, we get

$y+\log y+e^x{\cos }^{2}x=c$

Now $y\left(0\right)=1$

$\Rightarrow 1+0+e^0.1=c$

$\Rightarrow c=2$

$\therefore y+\log y+e^x{\cos }^{2}x=2$