MHT CET · Maths · Differential Equations
General solution of the differential equation \(\sin ^3 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dy}}=\sin \mathrm{y}\) is given by
- A \(\cos y-\frac{3}{4} \cos x-\frac{1}{12} \cos 3 x=C\)
- B \(\cos y-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x=C\)
- C \(\cos y+\frac{3}{4} \cos x-\frac{1}{12} \cos 3 x=C\)
- D \(\cos y+\frac{3}{4} \cos x-\frac{1}{12} \cos 3 x=C\)
Answer & Solution
Correct Answer
(B) \(\cos y-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x=C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \sin ^3 x \frac{d x}{d y}=\sin y \Rightarrow \int \sin ^3 x d x=\int \sin y d y \\ & \Rightarrow \int \frac{3 \sin x-\sin 3 x}{4} d x=\int \sin y d y \\ & \Rightarrow \frac{-3}{4} \cos x+\frac{1}{4} \cdot \frac{\cos 3 x}{3}=-\cos y+C \\ & \Rightarrow \cos y-\frac{3}{4} \cos x+\frac{1}{12} \cos 3 x=C\end{aligned}\)
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