Question

If general solution of the differential equation $e^{-x}(y+1)dy+({\cos }^{2}x-\sin (2x\left)\right)ydx=0$ is $y+\ln |y|+f\left(x\right)=c,$ then which of the following is/are true

• A. $f\left(x\right)=e^x{\cos }^{2}x$
• B. $f\left(x\right)=e^x\left(\sin 2x\right)$
• C. $c=2$ when $y=1,x=0$
  
• D. $c=1$ when $y=1,x=0$
  

Answer 1

Answer: (A), (C)

$c=2$ when $y=1,x=0$


$e^{-x}(y+1)dy+({\cos }^{2}x-\sin (2x\left)\right)ydx=0$
$\Rightarrow e^{-x}(y+1)dy=\left(\sin \right(2x)-{\cos }^{2}x)ydx$
$\Rightarrow \left(\frac{y+1}{y}\right)dy=e^x(\sin 2x-{\cos }^{2}x)dx$
Integrating both sides, we get
$\Rightarrow \int \left(\frac{y+1}{y}\right)dy=\int e^x(\sin 2x-{\cos }^{2}x)dx$
$\Rightarrow \int \left(\frac{y+1}{y}\right)dy=\int e^x\left(\sin 2x\right)dx-\int e^x\left({\cos }^{2}x\right)dx\cdots \left(i\right)$

Let $I_1=\int e^x\left({\cos }^{2}x\right)dx$
Using ILATE rule, we get
$I_1=e^x{\cos }^{2}x+\int e^x\left(\sin 2x\right)dx$
Putting in $\left(i\right),$ we get
$y+\ln |y|=-e^x{\cos }^{2}x+c$
$\Rightarrow y+\ln |y|+e^x{\cos }^{2}x=c$
When $y=1,x=0\Rightarrow c=2$