MHT CET · Maths · Differential Equations
The particular solution of the differential equation \(\frac{d y}{d x}=e^{2 y} \cos x\), when \(y\left(\frac{\pi}{6}\right)=0\) is
- A \(\sin \mathrm{x}-\frac{\mathrm{e}^{2 \mathrm{y}}}{2}=0\)
- B \(4 \sin x-e^{-2 y}-1=0\)
- C \(\sin x+e^{-2 y}-2=0\)
- D \(2 \sin x+e^{-2 y}-2=0\)
Answer & Solution
Correct Answer
(D) \(2 \sin x+e^{-2 y}-2=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{d y}{d x}=e^{2 y} \cos x \Rightarrow \int e^{2 y} d y=\int \cos d x \\ & \Rightarrow \frac{e^{2 y}}{-2}=\sin x+c\end{aligned}\)
Putting \(\mathrm{x}=\frac{\pi}{6}\) and \(\mathrm{y}=0\) we get \(\mathrm{c}=-1\)
\(\begin{aligned} & \Rightarrow \frac{e^{-2 y}}{-2}=\sin x-1 \\ & \Rightarrow e^{-2 y}=-2 \sin x+2 \\ & \Rightarrow 2 \sin x+e^{-2 y}-2=0\end{aligned}\)
Putting \(\mathrm{x}=\frac{\pi}{6}\) and \(\mathrm{y}=0\) we get \(\mathrm{c}=-1\)
\(\begin{aligned} & \Rightarrow \frac{e^{-2 y}}{-2}=\sin x-1 \\ & \Rightarrow e^{-2 y}=-2 \sin x+2 \\ & \Rightarrow 2 \sin x+e^{-2 y}-2=0\end{aligned}\)
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