MHT CET · Maths · Differential Equations
General solution of the differential equation \(\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0\) is
- A \((1+\cos x)(1+\sin y)=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- B \(1+\sin x+\cos y=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- C \((1+\sin x)(1+\cos y)=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
- D \(1+\sin x \cos y=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Answer & Solution
Correct Answer
(C) \((1+\sin x)(1+\cos y)=\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& \cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0 \\
& \Rightarrow \frac{\cos x}{1+\sin x} \mathrm{~d} x-\frac{\sin y}{1+\cos y} \mathrm{~d} y=0
\end{aligned}
\)
Integrating on both sides, we get
\(
\begin{aligned}
& \log |1+\sin x|+\log |1+\cos y|=\log |c| \\
& \Rightarrow \log |(1+\sin x)(1+\cos y)|=\log |c| \\
& \Rightarrow(1+\sin x)(1+\cos y)=c
\end{aligned}
\)
\begin{aligned}
& \cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0 \\
& \Rightarrow \frac{\cos x}{1+\sin x} \mathrm{~d} x-\frac{\sin y}{1+\cos y} \mathrm{~d} y=0
\end{aligned}
\)
Integrating on both sides, we get
\(
\begin{aligned}
& \log |1+\sin x|+\log |1+\cos y|=\log |c| \\
& \Rightarrow \log |(1+\sin x)(1+\cos y)|=\log |c| \\
& \Rightarrow(1+\sin x)(1+\cos y)=c
\end{aligned}
\)
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