Question

Differential Equations of first order and first Degree

Q. Cos(x+y) dy =dx

Answer 1

Dear student
$$\begin{gathered} \cos \left(x+y\right)dy=dx \\ \frac{dy}{dx}=\frac{1}{\cos \left(x+y\right)} \\ Letx+y=v.Then \\ 1+\frac{dy}{dx}=\frac{dv}{dx}\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1 \\ Puttingx+y=vand\frac{dy}{dx}=\frac{dv}{dx}-1thegivendifferentialeq,weget \\ \Rightarrow \frac{dv}{dx}-1=\frac{1}{\cos v} \\ \Rightarrow \frac{dv}{dx}=\frac{1+\cos v}{\cos v} \\ \Rightarrow \frac{\cos v}{1+\cos v}dv=dx \\ \Rightarrow \frac{\cos v\left(1-\cos v\right)}{1-{\cos }^{2}v}dv=dx \\ \Rightarrow \left(cotv\cos ecv-co{t}^{2}v\right)dv=dx \\ \Rightarrow \left(cotv\cos ecv-\cos e{c}^{2}v+1\right)dv=dx \\ \Rightarrow -\cos ecv+cotv+v=x+C\left[Onintegrating\right] \\ \Rightarrow -\cos ec\left(x+y\right)+cot\left(x+y\right)+y=C \\ \Rightarrow \frac{-1-\cos \left(x+y\right)}{\sin \left(x+y\right)}+y=C \\ \Rightarrow -\tan \left(\frac{x+y}{2}\right)+y=C \\ \Rightarrow y=\tan \left(\frac{x+y}{2}\right)+C \end{gathered}$$
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