The solution of the differential equation $\mathrm{e}^{-x}(y+1) \mathrm{d} y+\left(\cos ^2 x-\sin 2 x\right)…
The solution of the differential equation $\mathrm{e}^{-x}(y+1) \mathrm{d} y+\left(\cos ^2 x-\sin 2 x\right) y \mathrm{~d} x=0$ at $x=0$, $y=1$ is
- $(y+1)+\mathrm{e}^x \cos ^2 x=2$
- $y+\log y=\mathrm{e}^x \cos ^2 x$
- $\log (y+1)+\mathrm{e}^x \cos ^2 x=1$
- $y+\log y+\mathrm{e}^x \cos ^2 x=2$
Solution
$\begin{aligned}
& \mathrm{e}^{-x}(y+1) \mathrm{d} y+\left(\cos ^2 x-\sin 2 x\right) y \mathrm{~d} x=0 \\
& \Rightarrow \frac{y+1}{y} \mathrm{~d} y+\mathrm{e}^x\left(\cos ^2 x-\sin 2 x\right) \mathrm{d} x=0
\end{aligned}$
Integrating on both sides, we get
$\begin{aligned}
& \int\left(1+\frac{1}{y}\right) \mathrm{d} y+\int \mathrm{e}^x\left(\cos ^2 x-2 \sin x \cos x\right) \mathrm{d} x=\mathrm{c} \\
& \Rightarrow y+\log |y|+\mathrm{e}^x \cos ^2 x=\mathrm{c} \\
& \text { At } x=0, y=1 \\
\therefore \quad & 1+\log 1+\mathrm{e}^0 \cos ^2 0=\mathrm{c} \\
& \Rightarrow \mathrm{c}=2 \\
\therefore \quad & y+\log |y|+\mathrm{e}^x \cos ^2 x=2
\end{aligned}$
Asked in: MHT CET 2023 (13 May Shift 1)
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