General solution of the differential equation $\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d}…
General solution of the differential equation $\cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0$ is
- $(1+\cos x)(1+\sin y)=\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $1+\sin x+\cos y=\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $(1+\sin x)(1+\cos y)=\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
- $1+\sin x \cos y=\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
Solution
$\begin{aligned}
& \cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0 \\
& \Rightarrow \frac{\cos x}{1+\sin x} \mathrm{~d} x-\frac{\sin y}{1+\cos y} \mathrm{~d} y=0
\end{aligned}$
Integrating on both sides, we get
$\begin{aligned}
& \log |1+\sin x|+\log |1+\cos y|=\log |c| \\
& \Rightarrow \log |(1+\sin x)(1+\cos y)|=\log |c| \\
& \Rightarrow(1+\sin x)(1+\cos y)=c
\end{aligned}$
Asked in: MHT CET 2023 (14 May Shift 1)
Practice more Differential Equations questions on Aicharya