Q120 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Mathematics
The solution of the equation $\frac{d y}{d x}+2 y \tan x=\sin x$ satisfying $y=0$ when $x=\frac{\pi}{3}$, is
- Correct answerA$y=2 \sin ^2 x+\cos x-2$
- B$y=2 \sin ^2 x-\cos x-2$
- C$y=2 \cos ^2 x-\sin x+2$
- D$y=2 \cos x-\sin ^2 x-1$
Solution
Given, differential equation is
$$
\frac{d y}{d x}+2 y \tan x=\sin x
$$
Here, $P=2 \tan x$ and $Q=\sin x$
$$
\begin{array}{ll}
\therefore & I . F=e^{\int 2 \tan x d x} \\
& =e^{2 \ln |\sec x|}=e^{\ln \sec ^2 x}=\sec ^2 x
\end{array}
$$
Now, required solution is,
$$
\begin{aligned}
& y \times I \cdot F=\int Q \cdot I F d x+c \\
\Rightarrow \quad & y \times \sec ^2 x=\int \sin x \cdot \sec ^2 x d x+c \\
\Rightarrow \quad & y \sec ^2 x=\int \sec x \tan x+c \\
\Rightarrow \quad & y \sec ^2 x=\sec x+c
\end{aligned}
$$
Given, $y=0$ when $x=\frac{\pi}{3}$
$$
\begin{aligned}
& \Rightarrow \quad 0=2+c \\
& \Rightarrow \quad c=-2 \\
& \therefore y \sec ^2 x=\sec x-2
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & y=\frac{\sec x}{\sec ^2 x}-\frac{2}{\sec ^2 x} \\
\Rightarrow & y=\cos x-2 \cos ^2 x \\
\Rightarrow & y=\cos x-2\left(1-\sin ^2 x\right) \\
\Rightarrow & y=\cos x-2+2 \sin ^2 x \\
\Rightarrow & y=2 \sin ^2 x+\cos x-2
\end{array}
$$
Hence, the correct answer is option: 'A'.