The general solution of the differential equation $\cos (x+y) \frac{d y}{d x}=1$ is

The general solution of the differential equation $\cos (x+y) \frac{d y}{d x}=1$ is
  1. $y=\tan (x+y)+c$
  2. $y=\sec (x+y)+c$
  3. $y=\tan \left(\frac{x+y}{2}\right)+c$
  4. $y=\cot \left(\frac{x+y}{2}\right)+c$

Solution

$\cos (x+y) \frac{d y}{d x}=1$ Put $x+y=V \Rightarrow 1+\frac{d y}{d x}=\frac{d V}{d x}$ $\begin{aligned} & \therefore \cos \mathrm{V}\left(\frac{\mathrm{dV}}{\mathrm{dx}}-1\right)=1 \Rightarrow \cos \mathrm{V}\left(\frac{\mathrm{dV}}{\mathrm{dx}}\right)=1+\cos \mathrm{V} \\ & \therefore \int \frac{\cos \mathrm{V}}{1+\cos \mathrm{V}} \mathrm{dV}=\int \mathrm{dx} \\ & \therefore \int\left[\frac{1+\cos \mathrm{V}}{1+\cos \mathrm{V}}-\frac{1}{1+\cos \mathrm{V}}\right] \mathrm{dV} \\ & =\int \mathrm{dx} \Rightarrow \int \mathrm{dV}-\frac{1}{2} \int \sec ^2 \frac{\mathrm{V}}{2} \mathrm{dV}=\int \mathrm{dx} \end{aligned}$ $\begin{aligned} & \therefore \mathrm{V}-\frac{1}{2} \frac{\tan \left(\frac{\mathrm{V}}{2}\right)}{\left(\frac{1}{2}\right)}=\mathrm{x}+\mathrm{c} \Rightarrow \mathrm{V}-\tan \left(\frac{\mathrm{V}}{2}\right)=\mathrm{x}+\mathrm{c} \\ & \Rightarrow \mathrm{x}+\mathrm{y}-\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)=\mathrm{x}+\mathrm{c} \\ & \mathrm{y}=\tan \left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)+\mathrm{c} \end{aligned}$

Asked in: MHT CET 2021 (24 Sep Shift 1)

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