The solution of the differential equation $\cos (x+y) d y=d x$ given that $y(0)=0$ is

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Aicharya · Accepted Answer

Given differential equation $\cos (x+y) d y=d x \Rightarrow \frac{d y}{d x}=\sec (x+y)$ Put $x+y=t \Rightarrow 1+\frac{d y}{d x}=\frac{d t}{d x}$ $\begin{aligned} & \Rightarrow \quad \frac{d t}{d x}-1=\sec (t) \Rightarrow \int \frac{d t}{1+\sec t}=\int d x \ & \Rightarrow \quad \int \frac{\cos t}{1+\cot t} d t=\int d x \ & \Rightarrow \int\left(1-\frac{1}{1+\cos t}\right) d t=\int d x \ & \Rightarrow \quad \int\left[1-\frac{1}{2} \sec ^2\left(\frac{t}{2}\right)\right] d t=\int d x \ & \Rightarrow \quad t-\tan \left(\frac{t}{2}\right)=x+c \ & \Rightarrow \quad x+y-\tan \left(\frac{x+y}{2}\right)=x+c \ & \Rightarrow \quad y=\tan \left(\frac{x+y}{2}\right)+c \ & \because \quad f(0)=0 \Rightarrow c=0 \ & \therefore \quad y=\tan \left(\frac{x+y}{2}\right) \end{aligned}$

The solution of the differential equation \(\cos (x+y) d y=d x\) given that \(y(0)=0\) is

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