\displaystyle \text{Solve the following differential equations:}

\displaystyle \textbf{Question 1: }~\frac{dy}{dx}=(x+y+1)^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=(x+y+1)^{2}
\displaystyle \text{Putting } x+y+1=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting in the given equation,}
\displaystyle \frac{dv}{dx}-1=v^{2}
\displaystyle \Rightarrow \frac{dv}{dx}=v^{2}+1
\displaystyle \Rightarrow \frac{1}{v^{2}+1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{v^{2}+1}\,dv=\int dx
\displaystyle \Rightarrow \tan^{-1}v=x+C
\displaystyle \Rightarrow \tan^{-1}(x+y+1)=x+C

\displaystyle \textbf{Question 2: }~\frac{dy}{dx}\cos(x-y)=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}\cos(x-y)=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\cos(x-y)}
\displaystyle \text{Putting } x-y=v
\displaystyle \Rightarrow 1-\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=1-\frac{dv}{dx}
\displaystyle \text{Substituting, we get}
\displaystyle 1-\frac{dv}{dx}=\frac{1}{\cos v}
\displaystyle \Rightarrow \frac{dv}{dx}=1-\frac{1}{\cos v}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{\cos v-1}{\cos v}
\displaystyle \Rightarrow \frac{\cos v}{\cos v-1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{\cos v}{\cos v-1}\,dv=\int dx
\displaystyle \Rightarrow -\int \frac{\cos v(1+\cos v)}{1-\cos^{2}v}\,dv=\int dx
\displaystyle \Rightarrow -\int \frac{\cos v(1+\cos v)}{\sin^{2}v}\,dv=\int dx
\displaystyle \Rightarrow -\int (\cot v\,\mathrm{cosec}\,v+\cot^{2}v)\,dv=\int dx
\displaystyle \Rightarrow -\int (\cot v\,\mathrm{cosec}\,v+\mathrm{cosec}^{2}v-1)\,dv=\int dx
\displaystyle \Rightarrow -(-\mathrm{cosec}\,v-\cot v-v)=x+C
\displaystyle \Rightarrow \mathrm{cosec}(x-y)+\cot(x-y)+x-y=x+C
\displaystyle \Rightarrow \mathrm{cosec}(x-y)+\cot(x-y)-y=C
\displaystyle \Rightarrow \frac{1+\cos(x-y)}{\sin(x-y)}-y=C
\displaystyle \Rightarrow \cot\!\left(\frac{x-y}{2}\right)=y+C

\displaystyle \textbf{Question 3: }~\frac{dy}{dx}=\frac{(x-y)+3}{2(x-y)+5}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{(x-y)+3}{2(x-y)+5}
\displaystyle \text{Putting } x-y=v
\displaystyle \Rightarrow 1-\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=1-\frac{dv}{dx}
\displaystyle \text{Substituting, we get}
\displaystyle 1-\frac{dv}{dx}=\frac{v+3}{2v+5}
\displaystyle \Rightarrow \frac{dv}{dx}=1-\frac{v+3}{2v+5}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{2v+5-(v+3)}{2v+5}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{v+2}{2v+5}
\displaystyle \Rightarrow \frac{2v+5}{v+2}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{2v+5}{v+2}\,dv=\int dx
\displaystyle \Rightarrow \int \left(2+\frac{1}{v+2}\right)\,dv=\int dx
\displaystyle \Rightarrow 2v+\log|v+2|=x+C
\displaystyle \Rightarrow 2(x-y)+\log|x-y+2|=x+C

\displaystyle \textbf{Question 4: }~\frac{dy}{dx}=(x+y)^{2}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=(x+y)^{2}
\displaystyle \text{Let } x+y=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting in the given equation,}
\displaystyle \frac{dv}{dx}-1=v^{2}
\displaystyle \Rightarrow \frac{dv}{dx}=v^{2}+1
\displaystyle \Rightarrow \frac{1}{v^{2}+1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{1}{v^{2}+1}\,dv=\int dx
\displaystyle \Rightarrow \tan^{-1}v=x+C
\displaystyle \Rightarrow v=\tan(x+C)
\displaystyle \Rightarrow x+y=\tan(x+C)

\displaystyle \textbf{Question 5: }~(x+y)^{2}\frac{dy}{dx}=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x+y)^{2}\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{(x+y)^{2}}
\displaystyle \text{Let } x+y=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting in the given equation,}
\displaystyle \frac{dv}{dx}-1=\frac{1}{v^{2}}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{1}{v^{2}}+1
\displaystyle \Rightarrow \frac{v^{2}}{v^{2}+1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{v^{2}}{v^{2}+1}\,dv=\int dx
\displaystyle \Rightarrow \int \left(1-\frac{1}{v^{2}+1}\right)dv=\int dx
\displaystyle \Rightarrow v-\tan^{-1}v=x+C
\displaystyle \Rightarrow x+y-\tan^{-1}(x+y)=x+C
\displaystyle \Rightarrow y-\tan^{-1}(x+y)=C

\displaystyle \textbf{Question 6: }~\cos^{2}(x-2y)=1-2\frac{dy}{dx}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \cos^{2}(x-2y)=1-2\frac{dy}{dx}
\displaystyle \Rightarrow 2\frac{dy}{dx}=1-\cos^{2}(x-2y)
\displaystyle \Rightarrow 2\frac{dy}{dx}=\sin^{2}(x-2y)
\displaystyle \text{Let } x-2y=v
\displaystyle \Rightarrow \frac{dv}{dx}=1-2\frac{dy}{dx}
\displaystyle \Rightarrow 2\frac{dy}{dx}=1-\frac{dv}{dx}
\displaystyle \text{Substituting in } 2\frac{dy}{dx}=\sin^{2}(x-2y), \text{ we get}
\displaystyle 1-\frac{dv}{dx}=\sin^{2}v
\displaystyle \Rightarrow \frac{dv}{dx}=1-\sin^{2}v
\displaystyle \Rightarrow \frac{dv}{dx}=\cos^{2}v
\displaystyle \Rightarrow \sec^{2}v\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \sec^{2}v\,dv=\int dx
\displaystyle \Rightarrow \tan v=x+C
\displaystyle \Rightarrow \tan(x-2y)=x+C
\displaystyle \Rightarrow x=\tan(x-2y)-C

\displaystyle \textbf{Question 7: }~\frac{dy}{dx}=\sec(x+y).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\sec(x+y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{\cos(x+y)}
\displaystyle \text{Let } x+y=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting, we get}
\displaystyle \frac{dv}{dx}-1=\frac{1}{\cos v}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{\cos v+1}{\cos v}
\displaystyle \Rightarrow \frac{\cos v}{\cos v+1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{\cos v}{\cos v+1}\,dv=\int dx
\displaystyle \Rightarrow \int \frac{\cos v(1-\cos v)}{1-\cos^{2}v}\,dv=\int dx
\displaystyle \Rightarrow \int \frac{\cos v(1-\cos v)}{\sin^{2}v}\,dv=\int dx
\displaystyle \Rightarrow \int (\cot v\,\mathrm{cosec}\,v-\cot^{2}v)\,dv=\int dx
\displaystyle \Rightarrow \int (\cot v\,\mathrm{cosec}\,v-\mathrm{cosec}^{2}v+1)\,dv=\int dx
\displaystyle \Rightarrow -\mathrm{cosec}\,v+\cot v+v=x+C
\displaystyle \Rightarrow -\mathrm{cosec}(x+y)+\cot(x+y)+x+y=x+C
\displaystyle \Rightarrow -\mathrm{cosec}(x+y)+\cot(x+y)+y=C
\displaystyle \Rightarrow \frac{-1+\cos(x+y)}{\sin(x+y)}+y=C
\displaystyle \Rightarrow -\tan\!\left(\frac{x+y}{2}\right)+y=C
\displaystyle \Rightarrow y=\tan\!\left(\frac{x+y}{2}\right)+C

\displaystyle \textbf{Question 8: }~\frac{dy}{dx}=\tan(x+y).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\tan(x+y)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{\sin(x+y)}{\cos(x+y)}
\displaystyle \text{Let } x+y=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting, we get}
\displaystyle \frac{dv}{dx}-1=\frac{\sin v}{\cos v}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{\sin v+\cos v}{\cos v}
\displaystyle \Rightarrow \frac{\cos v}{\sin v+\cos v}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{\cos v}{\sin v+\cos v}\,dv=\int dx
\displaystyle \Rightarrow \frac{1}{2}\int \frac{(\sin v+\cos v)+(\cos v-\sin v)}{\sin v+\cos v}\,dv=\int dx
\displaystyle \Rightarrow \frac{1}{2}\int dv+\frac{1}{2}\int \frac{\cos v-\sin v}{\sin v+\cos v}\,dv=\int dx
\displaystyle \text{Let } \sin v+\cos v=t
\displaystyle \Rightarrow (\cos v-\sin v)\,dv=dt
\displaystyle \Rightarrow \frac{1}{2}v+\frac{1}{2}\log|t|=x+C
\displaystyle \Rightarrow \frac{1}{2}v+\frac{1}{2}\log|\sin v+\cos v|=x+C
\displaystyle \Rightarrow \frac{1}{2}(x+y)+\frac{1}{2}\log|\sin(x+y)+\cos(x+y)|=x+C
\displaystyle \Rightarrow y-x+\log|\sin(x+y)+\cos(x+y)|=C

\displaystyle \textbf{Question 9: }~(x+y)(dx-dy)=dx+dy.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x+y)(dx-dy)=dx+dy
\displaystyle \Rightarrow xdx+ydx-xdy-ydy=dx+dy
\displaystyle \Rightarrow (x+y-1)\,dx=(x+y+1)\,dy
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x+y-1}{x+y+1}
\displaystyle \text{Let } x+y=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting, we get}
\displaystyle \frac{dv}{dx}-1=\frac{v-1}{v+1}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{v-1}{v+1}+1
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{2v}{v+1}
\displaystyle \Rightarrow \frac{v+1}{2v}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{v+1}{2v}\,dv=\int dx
\displaystyle \Rightarrow \frac{1}{2}\int dv+\frac{1}{2}\int \frac{1}{v}\,dv=\int dx
\displaystyle \Rightarrow \frac{1}{2}v+\frac{1}{2}\log|v|=x+C
\displaystyle \Rightarrow \frac{1}{2}(x+y)+\frac{1}{2}\log|x+y|=x+C
\displaystyle \Rightarrow y-x+\log|x+y|=C

\displaystyle \textbf{Question 10: }~(x+y+1)\frac{dy}{dx}=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x+y+1)\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x+y+1}
\displaystyle \text{Let } x+y+1=v
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1
\displaystyle \text{Substituting, we get}
\displaystyle \frac{dv}{dx}-1=\frac{1}{v}
\displaystyle \Rightarrow \frac{dv}{dx}=\frac{1}{v}+1
\displaystyle \Rightarrow \frac{v}{v+1}\,dv=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int \frac{v}{v+1}\,dv=\int dx
\displaystyle \Rightarrow \int \left(1-\frac{1}{v+1}\right)dv=\int dx
\displaystyle \Rightarrow v-\log|v+1|=x+C
\displaystyle \Rightarrow x+y+1-\log|x+y+2|=x+C
\displaystyle \Rightarrow y-\log|x+y+2|=C
\displaystyle \Rightarrow \log|x+y+2|=y-C
\displaystyle \Rightarrow |x+y+2|=e^{\,y-C}
\displaystyle \Rightarrow x+y+2=Ce^{\,y}
\displaystyle \Rightarrow x=Ce^{\,y}-y-2

\displaystyle \textbf{Question 11: }~\frac{dy}{dx}+1=e^{x+y}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}+1=e^{x+y}\quad (1)
\displaystyle \text{Let } x+y=t
\displaystyle \Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}
\displaystyle \text{Substituting in (1), we get}
\displaystyle \frac{dt}{dx}=e^{t}
\displaystyle \Rightarrow e^{-t}\,dt=dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \int e^{-t}\,dt=\int dx
\displaystyle \Rightarrow -e^{-t}=x+C
\displaystyle \Rightarrow -e^{-(x+y)}=x+C

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