Q120 / 180 · COMEDK 2016 · Mathematics
The solution of the differential equation $e^{-x}(y+1) d y+\left(\cos ^{2} x-\sin 2 x\right) y(d x)=0$ subjected to the condition that $y=1$, when $x=0$ is
- A$(y+1)+e^{x} \cos ^{2} x=2$
- B$y+\log y=e^{x} \cos ^{2} x$
- C$\log (y+1)+e^{x} \cos ^{2} x=1$
- Correct answerD$y+\log y+e^{x} \cos ^{2} x=2$
Solution
Given, differential equation is
$\begin{aligned}
& e^{-x}(y+1) d y+\left(\cos ^{2} x-\sin 2 x\right) y d x=0 \\
\Rightarrow &\left(\frac{y+1}{y}\right) d y+e^{x}\left(\cos ^{2} x-\sin 2 x\right) d x=0 \\
\Rightarrow &\left(1+\frac{1}{y}\right) d y+\left(e^{x} \cos ^{2} x-e^{x} \sin 2 x\right) d x=0 \\
\Rightarrow & d(y+\log y)+d\left(e^{x} \cos ^{2} x\right)=0
\end{aligned}$
On integrating above equation, we get
$\begin{aligned}
&y+\log y+e^{x} \cos ^{2} x=c \\
&\text { At } y(0)=1 \\
&\Rightarrow 1+0+e^{0} \cdot 1=c \Rightarrow c=2
\end{aligned}$
Hence, required solution is
$y+\log y+e^{x} \cos ^{2} x=2$
Hence, the correct answer is option: 'D'.