Q92 / 150 · MHT CET 2023 (10 May Shift 1) · Mathematics

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
    $(1+\cos x)(1+\sin y)=\mathrm{c}$
  • B
    $1+\sin x+\cos y=c$
  • Correct answer
    C
    $(1+\sin x)(1+\cos y)=c$
  • D
    $1+\sin x \cdot \cos y=c$

Solution

Given differential equation is $\begin{aligned} & \cos x(1+\cos y) \mathrm{d} x-\sin y(1+\sin x) \mathrm{d} y=0 \\ & \Rightarrow \cos x(1+\cos y) \mathrm{d} x=\sin y(1+\sin x) \mathrm{d} y \\ & \Rightarrow \frac{\cos x}{1+\sin x} \mathrm{~d} x=\frac{\sin y}{1+\cos y} \mathrm{~d} y \end{aligned}$ Integrating on both sides, we get $\begin{aligned} & \int \frac{\cos x}{1+\sin x} \mathrm{~d} x=\frac{\sin y}{1+\cos y} \mathrm{~d} y+\log |\mathrm{c}| \\ & \Rightarrow \log |1+\sin x|=-\log |1+\cos y|+\log |\mathrm{c}| \\ & \Rightarrow \log |1+\sin x|+\log |1+\cos y|=\log |\mathrm{c}| \\ & \Rightarrow \log |(1+\sin x)(1+\cos y)|=\log |\mathrm{c}| \\ & \Rightarrow(1+\sin x)(1+\cos y)=\mathrm{c} \end{aligned}$

Hence, the correct answer is option: 'C'.

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