MHT CET · Maths · Differential Equations
The general solution of the differential equation \(\cos x \cdot \sin y d x\) \(+\sin x \cdot \cos y d y=0\) is
- A \(\sin x+\sin y=c\)
- B \(\cos x+\cos y=c\)
- C \(\sin x \cdot \sin y=c\)
- D \(\cos x \cdot \cos y=c\)
Answer & Solution
Correct Answer
(C) \(\sin x \cdot \sin y=c\)
Step-by-step Solution
Detailed explanation
\( \cos \mathrm{x} \sin \mathrm{ydx}+\sin \mathrm{x} \cos \mathrm{y} d \mathrm{y}=0 \)
\( \therefore \cos \mathrm{x} \sin \mathrm{ydx}=-\sin \mathrm{x} \cos \mathrm{ydy} \)
\( \therefore \int \frac{\cos \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=-\int \frac{\cos \mathrm{y}}{\sin \mathrm{y}} \mathrm{dy} \)
\( \therefore \log |\sin \mathrm{x}|=-\log |\sin \mathrm{y}|+\mathrm{c}_1 \)
\( \therefore \log |\sin \mathrm{x}|+\log |\sin \mathrm{y}|=\mathrm{c}_1 \Rightarrow\) \(\log [\sin \mathrm{x} \sin \mathrm{y}]=\mathrm{c}_1 \)
\( \therefore \sin \mathrm{x} \sin \mathrm{y}=\mathrm{e}^{c_1}=\mathrm{c}\)
\( \therefore \cos \mathrm{x} \sin \mathrm{ydx}=-\sin \mathrm{x} \cos \mathrm{ydy} \)
\( \therefore \int \frac{\cos \mathrm{x}}{\sin \mathrm{x}} \mathrm{dx}=-\int \frac{\cos \mathrm{y}}{\sin \mathrm{y}} \mathrm{dy} \)
\( \therefore \log |\sin \mathrm{x}|=-\log |\sin \mathrm{y}|+\mathrm{c}_1 \)
\( \therefore \log |\sin \mathrm{x}|+\log |\sin \mathrm{y}|=\mathrm{c}_1 \Rightarrow\) \(\log [\sin \mathrm{x} \sin \mathrm{y}]=\mathrm{c}_1 \)
\( \therefore \sin \mathrm{x} \sin \mathrm{y}=\mathrm{e}^{c_1}=\mathrm{c}\)
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