\displaystyle \text{Solve the following differential equations (1-21):}

\displaystyle \textbf{Question 1: }~\frac{dy}{dx}=x^{2}+x-\frac{1}{x},\ x\neq 0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=x^{2}+x-\frac{1}{x}
\displaystyle \Rightarrow dy=\left(x^{2}+x-\frac{1}{x}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(x^{2}+x-\frac{1}{x}\right)dx
\displaystyle \Rightarrow y=\frac{x^{3}}{3}+\frac{x^{2}}{2}-\log|x|+C
\displaystyle \text{Clearly, } y=\frac{x^{3}}{3}+\frac{x^{2}}{2}-\log|x|+C \text{ is defined for all } x\in R \text{ except } x=0
\displaystyle \text{Hence, } y=\frac{x^{3}}{3}+\frac{x^{2}}{2}-\log|x|+C,\ \text{where } x\in R-\{0\}, \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 2: }~\frac{dy}{dx}=x^{5}+x^{2}-\frac{2}{x},\ x\neq 0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=x^{5}+x^{2}-\frac{2}{x}
\displaystyle \Rightarrow dy=\left(x^{5}+x^{2}-\frac{2}{x}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(x^{5}+x^{2}-\frac{2}{x}\right)dx
\displaystyle \Rightarrow y=\frac{x^{6}}{6}+\frac{x^{3}}{3}-2\log|x|+C
\displaystyle \text{Clearly, } y=\frac{x^{6}}{6}+\frac{x^{3}}{3}-2\log|x|+C \text{ is defined for all } x\in R \text{ except } x=0
\displaystyle \text{Hence, } y=\frac{x^{6}}{6}+\frac{x^{3}}{3}-2\log|x|+C,\ \text{where } x\in R-\{0\}, \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 3: }~\frac{dy}{dx}+2x=e^{3x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}+2x=e^{3x}
\displaystyle \Rightarrow \frac{dy}{dx}=e^{3x}-2x
\displaystyle \Rightarrow dy=(e^{3x}-2x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(e^{3x}-2x)\,dx
\displaystyle \Rightarrow y=\frac{e^{3x}}{3}-\frac{2x^{2}}{2}+C
\displaystyle \Rightarrow y=\frac{e^{3x}}{3}-x^{2}+C
\displaystyle \Rightarrow y+x^{2}=\frac{e^{3x}}{3}+C
\displaystyle \text{So, } y+x^{2}=\frac{e^{3x}}{3}+C \text{ is defined for all } x\in R
\displaystyle \text{Hence, } y+x^{2}=\frac{e^{3x}}{3}+C,\ \text{where } x\in R,\ \text{is the solution to the given differential equation.}

\displaystyle \textbf{Question 4: }~(x^{2}+1)\frac{dy}{dx}=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{2}+1)\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x^{2}+1}
\displaystyle \Rightarrow dy=\left(\frac{1}{x^{2}+1}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(\frac{1}{x^{2}+1}\right)dx
\displaystyle \Rightarrow y=\tan^{-1}x+C
\displaystyle \text{So, } y=\tan^{-1}x+C \text{ is defined for all } x\in R
\displaystyle \text{Hence, } y=\tan^{-1}x+C,\ \text{where } x\in R,\ \text{is the solution to the given differential equation.}

\displaystyle \textbf{Question 5: }~\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2\sin^{2}\frac{x}{2}}{2\cos^{2}\frac{x}{2}}
\displaystyle \Rightarrow \frac{dy}{dx}=\tan^{2}\frac{x}{2}
\displaystyle \Rightarrow dy=\left(\tan^{2}\frac{x}{2}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(\tan^{2}\frac{x}{2}\right)dx
\displaystyle \Rightarrow \int dy=\int\left(\sec^{2}\frac{x}{2}-1\right)dx
\displaystyle \Rightarrow y=2\tan\frac{x}{2}-x+C
\displaystyle \text{So, } y=2\tan\frac{x}{2}-x+C \text{ is defined for all } x\in R
\displaystyle \text{Hence, } y=2\tan\frac{x}{2}-x+C,\ \text{where } x\in R,\ \text{is the solution to the given differential equation.}

\displaystyle \textbf{Question 6: }~(x+2)\frac{dy}{dx}=x^{2}+3x+7.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x+2)\frac{dy}{dx}=x^{2}+3x+7
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{x^{2}+3x+7}{x+2}
\displaystyle \Rightarrow dy=\left(\frac{x^{2}+3x+7}{x+2}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(\frac{x^{2}+3x+7}{x+2}\right)dx
\displaystyle \Rightarrow \int dy=\int\left(\frac{x^{2}+3x+2+5}{x+2}\right)dx
\displaystyle \Rightarrow \int dy=\int\left(\frac{(x+2)(x+1)+5}{x+2}\right)dx
\displaystyle \Rightarrow \int dy=\int\left(x+1+\frac{5}{x+2}\right)dx
\displaystyle \Rightarrow y=\frac{x^{2}}{2}+x+5\log|x+2|+C
\displaystyle \text{So, } y=\frac{x^{2}}{2}+x+5\log|x+2|+C \text{ is defined for all } x\in R \text{ except } x=-2
\displaystyle \text{Hence, } y=\frac{x^{2}}{2}+x+5\log|x+2|+C,\ \text{where } x\in R-\{-2\},  \\ \text{is the solution to the given differential equation.}

\displaystyle \textbf{Question 7: }~\frac{dy}{dx}=\tan^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\tan^{-1}x
\displaystyle \Rightarrow dy=(\tan^{-1}x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(\tan^{-1}x)\,dx
\displaystyle \Rightarrow y=\int 1\cdot \tan^{-1}x\,dx
\displaystyle \Rightarrow y=\tan^{-1}x\int 1\,dx-\int\left[\frac{d}{dx}(\tan^{-1}x)\int 1\,dx\right]dx
\displaystyle \Rightarrow y=x\tan^{-1}x-\int\frac{x}{1+x^{2}}\,dx
\displaystyle \Rightarrow y=x\tan^{-1}x-\frac{1}{2}\int\frac{2x}{1+x^{2}}\,dx
\displaystyle \Rightarrow y=x\tan^{-1}x-\frac{1}{2}\log|1+x^{2}|+C
\displaystyle \text{So, } y=x\tan^{-1}x-\frac{1}{2}\log|1+x^{2}|+C \text{ is defined for all } x\in R
\displaystyle \text{Hence, } y=x\tan^{-1}x-\frac{1}{2}\log|1+x^{2}|+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 8: }~\frac{dy}{dx}=\log x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\log x
\displaystyle \Rightarrow dy=(\log x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(\log x)\,dx
\displaystyle \Rightarrow y=\int 1\cdot \log x\,dx
\displaystyle \Rightarrow y=\log x\int 1\,dx-\int\left[\frac{d}{dx}(\log x)\int 1\,dx\right]dx
\displaystyle \Rightarrow y=x\log x-\int\frac{x}{x}\,dx
\displaystyle \Rightarrow y=x\log x-\int 1\,dx
\displaystyle \Rightarrow y=x\log x-x
\displaystyle \Rightarrow y=x(\log x-1)+C
\displaystyle \text{So, } y=x(\log x-1)+C \text{ is defined for all } x\in R \text{ except } x=0
\displaystyle \text{Hence, } y=x(\log x-1)+C,\ \text{where } x\in R-\{0\}, \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 9: }~\frac{1}{x}\frac{dy}{dx}=\tan^{-1}x,\ x\neq 0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{1}{x}\frac{dy}{dx}=\tan^{-1}x
\displaystyle \Rightarrow \frac{dy}{dx}=x\tan^{-1}x
\displaystyle \Rightarrow dy=(x\tan^{-1}x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(x\tan^{-1}x)\,dx
\displaystyle \Rightarrow y=\int x\tan^{-1}x\,dx
\displaystyle \Rightarrow y=\tan^{-1}x\int x\,dx-\int\left[\frac{d}{dx}(\tan^{-1}x)\int x\,dx\right]dx
\displaystyle \Rightarrow y=\frac{x^{2}\tan^{-1}x}{2}-\frac{1}{2}\int\frac{x^{2}}{1+x^{2}}\,dx
\displaystyle \Rightarrow y=\frac{x^{2}\tan^{-1}x}{2}-\frac{1}{2}\int\frac{x^{2}+1-1}{1+x^{2}}\,dx
\displaystyle \Rightarrow y=\frac{x^{2}\tan^{-1}x}{2}-\frac{1}{2}\int\left(1-\frac{1}{1+x^{2}}\right)dx
\displaystyle \Rightarrow y=\frac{x^{2}\tan^{-1}x}{2}-\frac{1}{2}x+\frac{1}{2}\tan^{-1}x+C
\displaystyle \Rightarrow y=\frac{(x^{2}+1)\tan^{-1}x}{2}-\frac{1}{2}x+C
\displaystyle \text{Hence, } y=\frac{(x^{2}+1)\tan^{-1}x}{2}-\frac{1}{2}x+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 10: }~\frac{dy}{dx}=\cos^{3}x\,\sin^{2}x+x\sqrt{2x+1}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=\cos^{3}x\sin^{2}x+x\sqrt{2x+1}
\displaystyle \Rightarrow dy=(\cos^{3}x\sin^{2}x+x\sqrt{2x+1})\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(\cos^{3}x\sin^{2}x+x\sqrt{2x+1})\,dx
\displaystyle \Rightarrow y=\int\cos^{3}x\sin^{2}x\,dx+\int x\sqrt{2x+1}\,dx
\displaystyle \Rightarrow y=I_{1}+I_{2} \qquad (1)
\displaystyle \text{where}
\displaystyle I_{1}=\int\cos^{3}x\sin^{2}x\,dx
\displaystyle I_{2}=\int x\sqrt{2x+1}\,dx
\displaystyle \text{Now,}
\displaystyle I_{1}=\int\cos^{3}x\sin^{2}x\,dx
\displaystyle =\int\sin^{2}x(1-\sin^{2}x)\cos x\,dx
\displaystyle \text{Putting } t=\sin x,\ \text{we get}
\displaystyle dt=\cos x\,dx
\displaystyle \Rightarrow I_{1}=\int t^{2}(1-t^{2})\,dt
\displaystyle =\int(t^{2}-t^{4})\,dt
\displaystyle =\frac{t^{3}}{3}-\frac{t^{5}}{5}+C_{1}
\displaystyle =\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C_{1}
\displaystyle I_{2}=\int x\sqrt{2x+1}\,dx
\displaystyle \text{Putting } t^{2}=2x+1,\ \text{we get}
\displaystyle 2t\,dt=2\,dx
\displaystyle \Rightarrow t\,dt=dx
\displaystyle \text{Now,}
\displaystyle I_{2}=\int\left(\frac{t^{2}-1}{2}\right)t\cdot t\,dt
\displaystyle =\frac{1}{2}\int(t^{4}-t^{2})\,dt
\displaystyle =\frac{t^{5}}{10}-\frac{t^{3}}{6}+C_{2}
\displaystyle =\frac{(2x+1)^{5/2}}{10}-\frac{(2x+1)^{3/2}}{6}+C_{2}
\displaystyle \text{Putting the values of } I_{1} \text{ and } I_{2} \text{ in (1), we get}
\displaystyle y=\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+C_{1}+\frac{(2x+1)^{5/2}}{10}-\frac{(2x+1)^{3/2}}{6}+C_{2}
\displaystyle \Rightarrow y=\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+\frac{(2x+1)^{5/2}}{10}-\frac{(2x+1)^{3/2}}{6}+C \qquad (\text{where } C=C_{1}+C_{2})
\displaystyle \text{Hence, } y=\frac{\sin^{3}x}{3}-\frac{\sin^{5}x}{5}+\frac{(2x+1)^{5/2}}{10}-\frac{(2x+1)^{3/2}}{6}+C \\ \text{is the solution to the given differential equation.}

\displaystyle \textbf{Question 11: }~(\sin x+\cos x)\,dy+(\cos x-\sin x)\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (\sin x+\cos x)\,dy+(\cos x-\sin x)\,dx=0
\displaystyle \Rightarrow dy=-\left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=-\int\left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx
\displaystyle \Rightarrow y=-\int\left(\frac{\cos x-\sin x}{\sin x+\cos x}\right)dx
\displaystyle \text{Putting } \sin x+\cos x=t
\displaystyle \Rightarrow (\cos x-\sin x)\,dx=dt
\displaystyle \therefore y=-\int\frac{dt}{t}
\displaystyle \Rightarrow y=-\log|t|+C
\displaystyle \Rightarrow y=-\log|\sin x+\cos x|+C
\displaystyle \Rightarrow y+\log|\sin x+\cos x|=C
\displaystyle \text{Hence, } y+\log|\sin x+\cos x|=C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 12: }~\frac{dy}{dx}-x\sin^{2}x=\frac{1}{x\log x}.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}-x\sin^{2}x=\frac{1}{x\log x}
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+x\sin^{2}x
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+\frac{x}{2}(1-\cos 2x)
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x
\displaystyle \Rightarrow dy=\left[\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x\right]dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left[\frac{1}{x\log x}+\frac{x}{2}-\frac{x}{2}\cos 2x\right]dx
\displaystyle \Rightarrow y=\int\frac{1}{x\log x}\,dx+\frac{1}{2}\int x\,dx-\frac{1}{2}\int x\cos 2x\,dx
\displaystyle \Rightarrow y=\log|\log x|+\frac{x^{2}}{4}-\frac{1}{2}\int x\cos 2x\,dx
\displaystyle \Rightarrow y=\log|\log x|+\frac{x^{2}}{4}-\frac{1}{2}\left[x\int\cos 2x\,dx-\int\frac{d}{dx}(x)\int\cos 2x\,dx\,dx\right]
\displaystyle \Rightarrow y=\log|\log x|+\frac{x^{2}}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C
\displaystyle \text{Hence, } y=\log|\log x|+\frac{x^{2}}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 13: }~\frac{dy}{dx}=x^{5}\tan^{-1}(x^{3}).
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=x^{5}\tan^{-1}(x^{3})
\displaystyle \Rightarrow dy=\{x^{5}\tan^{-1}(x^{3})\}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int x^{5}\tan^{-1}(x^{3})\,dx
\displaystyle \Rightarrow y=\int x^{5}\tan^{-1}(x^{3})\,dx
\displaystyle \text{Putting } t=x^{3}, \text{ we get}
\displaystyle dt=3x^{2}\,dx
\displaystyle \Rightarrow y=\frac{1}{3}\int t\tan^{-1}t\,dt
\displaystyle \Rightarrow y=\frac{1}{3}\int t\cdot \tan^{-1}t\,dt
\displaystyle \Rightarrow y=\frac{1}{3}\left[\tan^{-1}t\int t\,dt-\int\left\{\frac{d}{dt}(\tan^{-1}t)\int t\,dt\right\}dt\right]
\displaystyle \Rightarrow y=\frac{1}{3}\left[\frac{t^{2}\tan^{-1}t}{2}-\int\frac{t^{2}}{2(1+t^{2})}\,dt\right]
\displaystyle \Rightarrow y=\frac{t^{2}\tan^{-1}t}{6}-\frac{1}{6}\int\frac{t^{2}+1-1}{1+t^{2}}\,dt
\displaystyle \Rightarrow y=\frac{t^{2}\tan^{-1}t}{6}-\frac{1}{6}\int dt+\frac{1}{6}\int\frac{1}{1+t^{2}}\,dt
\displaystyle \Rightarrow y=\frac{t^{2}\tan^{-1}t}{6}-\frac{t}{6}+\frac{\tan^{-1}t}{6}+C
\displaystyle \Rightarrow y=\frac{x^{6}\tan^{-1}(x^{3})}{6}-\frac{x^{3}}{6}+\frac{\tan^{-1}(x^{3})}{6}+C
\displaystyle \Rightarrow y=\frac{1}{6}\left(x^{6}\tan^{-1}(x^{3})-x^{3}+\tan^{-1}(x^{3})\right)+C
\displaystyle \text{Hence, } y=\frac{1}{6}\left(x^{6}\tan^{-1}(x^{3})-x^{3}+\tan^{-1}(x^{3})\right)+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 14: }~\sin^{4}x\,\frac{dy}{dx}=\cos x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \sin^{4}x\,\frac{dy}{dx}=\cos x
\displaystyle \Rightarrow dy=\frac{\cos x}{\sin^{4}x}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\frac{\cos x}{\sin^{4}x}\,dx
\displaystyle \Rightarrow y=\int\frac{\cos x}{\sin^{4}x}\,dx
\displaystyle \text{Putting } \sin x=t
\displaystyle \Rightarrow \cos x\,dx=dt
\displaystyle \therefore y=\int\frac{1}{t^{4}}\,dt
\displaystyle \Rightarrow y=\frac{t^{-3}}{-3}+C
\displaystyle \Rightarrow y=-\frac{1}{3}\sin^{-3}x+C
\displaystyle \Rightarrow y=-\frac{1}{3}\mathrm{cosec}^{3}x+C
\displaystyle \text{Hence, } y=-\frac{1}{3}\mathrm{cosec}^{3}x+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 15: }~\cos x\,\frac{dy}{dx}-\cos 2x=\cos 3x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \cos x\,\frac{dy}{dx}-\cos 2x=\cos 3x
\displaystyle \Rightarrow \cos x\,\frac{dy}{dx}=\cos 3x+\cos 2x
\displaystyle \Rightarrow dy=\frac{\cos 3x+\cos 2x}{\cos x}\,dx
\displaystyle \Rightarrow dy=\frac{4\cos^{3}x-3\cos x+2\cos^{2}x-1}{\cos x}\,dx
\displaystyle \Rightarrow dy=(4\cos^{2}x-3+2\cos x-\sec x)\,dx
\displaystyle \Rightarrow dy=[2(2\cos^{2}x-1)-1+2\cos x-\sec x]\,dx
\displaystyle \Rightarrow dy=(2\cos 2x-1+2\cos x-\sec x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(2\cos 2x-1+2\cos x-\sec x)\,dx
\displaystyle \Rightarrow y=\sin 2x-x+2\sin x-\log|\sec x+\tan x|+C
\displaystyle \text{Hence, } y=\sin 2x-x+2\sin x-\log|\sec x+\tan x|+C \\ \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 16: }~\sqrt{1-x^{4}}\,dy=x\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \sqrt{1-x^{4}}\,dy=x\,dx
\displaystyle \Rightarrow dy=\frac{x}{\sqrt{1-x^{4}}}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\frac{x}{\sqrt{1-x^{4}}}\,dx
\displaystyle \Rightarrow y=\int\frac{x}{\sqrt{1-x^{4}}}\,dx
\displaystyle \text{Putting } x^{2}=t
\displaystyle \Rightarrow 2x\,dx=dt
\displaystyle \therefore y=\frac{1}{2}\int\frac{dt}{\sqrt{1-t^{2}}}
\displaystyle \Rightarrow y=\frac{1}{2}\sin^{-1}t+C
\displaystyle \Rightarrow y=\frac{1}{2}\sin^{-1}(x^{2})+C
\displaystyle \text{Hence, } y=\frac{1}{2}\sin^{-1}(x^{2})+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 17: }~\sqrt{a+x}\,dy+x\,dx=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \sqrt{a+x}\,dy+x\,dx=0
\displaystyle \Rightarrow \sqrt{a+x}\,dy=-x\,dx
\displaystyle \Rightarrow dy=-\frac{x}{\sqrt{a+x}}\,dx
\displaystyle \Rightarrow dy=-\frac{x+a-a}{\sqrt{a+x}}\,dx
\displaystyle \Rightarrow dy=-\left(\sqrt{a+x}-\frac{a}{\sqrt{a+x}}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=-\int\left(\sqrt{a+x}-\frac{a}{\sqrt{a+x}}\right)dx
\displaystyle \Rightarrow y=-\frac{2}{3}(a+x)^{3/2}+2a\sqrt{a+x}+C
\displaystyle \Rightarrow y+\frac{2}{3}(a+x)^{3/2}-2a\sqrt{a+x}=C
\displaystyle \text{Hence, } y+\frac{2}{3}(a+x)^{3/2}-2a\sqrt{a+x}=C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 18: }~(1+x^{2})\frac{dy}{dx}-x=2\tan^{-1}x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (1+x^{2})\frac{dy}{dx}-x=2\tan^{-1}x
\displaystyle \Rightarrow (1+x^{2})\frac{dy}{dx}=x+2\tan^{-1}x
\displaystyle \Rightarrow dy=\left\{\frac{x}{1+x^{2}}+\frac{2}{1+x^{2}}\tan^{-1}x\right\}dx
\displaystyle \Rightarrow dy=\left\{\frac{1}{2}\cdot\frac{2x}{1+x^{2}}+\frac{2}{1+x^{2}}\tan^{-1}x\right\}dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left\{\frac{1}{2}\cdot\frac{2x}{1+x^{2}}+\frac{2}{1+x^{2}}\tan^{-1}x\right\}dx
\displaystyle \Rightarrow y=\frac{1}{2}\int\frac{2x}{1+x^{2}}\,dx+2\int\frac{1}{1+x^{2}}\tan^{-1}x\,dx
\displaystyle \Rightarrow y=\frac{1}{2}\log|1+x^{2}|+2\int\frac{1}{1+x^{2}}\tan^{-1}x\,dx
\displaystyle \text{Putting } \tan^{-1}x=t
\displaystyle \Rightarrow \frac{1}{1+x^{2}}\,dx=dt
\displaystyle \therefore y=\frac{1}{2}\log|1+x^{2}|+2\int t\,dt
\displaystyle \Rightarrow y=\frac{1}{2}\log|1+x^{2}|+t^{2}+C
\displaystyle \Rightarrow y=\frac{1}{2}\log|1+x^{2}|+(\tan^{-1}x)^{2}+C
\displaystyle \text{Hence, } y=\frac{1}{2}\log|1+x^{2}|+(\tan^{-1}x)^{2}+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 19: }~\frac{dy}{dx}=x\log x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=x\log x
\displaystyle \Rightarrow dy=(x\log x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(x\log x)\,dx
\displaystyle \Rightarrow y=\int x\log x\,dx
\displaystyle \Rightarrow y=\log x\int x\,dx-\int\left[\frac{d}{dx}(\log x)\int x\,dx\right]dx
\displaystyle \Rightarrow y=\log x\cdot\frac{x^{2}}{2}-\int\left(\frac{1}{x}\cdot\frac{x^{2}}{2}\right)dx
\displaystyle \Rightarrow y=\frac{x^{2}}{2}\log x-\int\frac{x}{2}\,dx
\displaystyle \Rightarrow y=\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}+C
\displaystyle \text{Hence, } y=\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 20: }~\frac{dy}{dx}=xe^{x}-\frac{5}{2}+\cos^{2}x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \frac{dy}{dx}=x e^{x}-\frac{5}{2}+\cos^{2}x
\displaystyle \Rightarrow \frac{dy}{dx}=x e^{x}-\frac{5}{2}+\frac{\cos 2x}{2}+\frac{1}{2}
\displaystyle \Rightarrow \frac{dy}{dx}=x e^{x}+\frac{\cos 2x}{2}-2
\displaystyle \Rightarrow dy=\left(x e^{x}+\frac{\cos 2x}{2}-2\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(x e^{x}+\frac{\cos 2x}{2}-2\right)dx
\displaystyle \Rightarrow y=\int x e^{x}dx+\frac{1}{2}\int\cos 2x\,dx-2\int dx
\displaystyle \Rightarrow y=x\int e^{x}dx-\int\left[\frac{d}{dx}(x)\int e^{x}dx\right]dx+\frac{1}{2}\cdot\frac{\sin 2x}{2}-2x
\displaystyle \Rightarrow y=x e^{x}-\int e^{x}dx+\frac{1}{4}\sin 2x-2x
\displaystyle \Rightarrow y=x e^{x}-e^{x}+\frac{1}{4}\sin 2x-2x+C
\displaystyle \text{Hence, } y=x e^{x}-e^{x}+\frac{1}{4}\sin 2x-2x+C \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 21: }~(x^{3}+x^{2}+x+1)\frac{dy}{dx}=2x^{2}+x.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle (x^{3}+x^{2}+x+1)\frac{dy}{dx}=2x^{2}+x
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{2x^{2}+x}{x^{3}+x^{2}+x+1}
\displaystyle \Rightarrow dy=\frac{2x^{2}+x}{(x+1)(x^{2}+1)}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left\{\frac{2x^{2}+x}{(x+1)(x^{2}+1)}\right\}dx
\displaystyle \Rightarrow y=\int\left\{\frac{2x^{2}+x}{(x+1)(x^{2}+1)}\right\}dx
\displaystyle \text{Let } \frac{2x^{2}+x}{(x+1)(x^{2}+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+1}
\displaystyle \Rightarrow 2x^{2}+x=A(x^{2}+1)+(Bx+C)(x+1)
\displaystyle \Rightarrow 2x^{2}+x=(A+B)x^{2}+(B+C)x+(A+C)
\displaystyle \text{Comparing the coefficients on both sides, we get}
\displaystyle A+B=2 \qquad (1)
\displaystyle B+C=1 \qquad (2)
\displaystyle A+C=0 \qquad (3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle A=\frac{1}{2},\ B=\frac{3}{2},\ C=-\frac{1}{2}
\displaystyle \Rightarrow y=\frac{1}{2}\int\frac{1}{x+1}\,dx+\int\frac{\frac{3}{2}x-\frac{1}{2}}{x^{2}+1}\,dx
\displaystyle \Rightarrow y=\frac{1}{2}\int\frac{1}{x+1}\,dx+\frac{1}{2}\int\frac{3x}{x^{2}+1}\,dx-\frac{1}{2}\int\frac{1}{x^{2}+1}\,dx
\displaystyle \Rightarrow y=\frac{1}{2}\log|x+1|+\frac{3}{4}\log|x^{2}+1|-\frac{1}{2}\tan^{-1}x+C
\displaystyle \text{Hence, } y=\frac{1}{2}\log|x+1|+\frac{3}{4}\log|x^{2}+1|-\frac{1}{2}\tan^{-1}x+C \\ \text{is the solution to the given differential equation.}

\displaystyle \text{Solve the following initial value problems (22-26):}

\displaystyle \textbf{Question 22: }~\sin\left(\frac{dy}{dx}\right)=k;\ y(0)=1.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle \sin\!\left(\frac{dy}{dx}\right)=k
\displaystyle \Rightarrow \frac{dy}{dx}=\sin^{-1}k
\displaystyle \Rightarrow dy=\{\sin^{-1}k\}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int(\sin^{-1}k)\,dx
\displaystyle \Rightarrow y=x\sin^{-1}k+C \qquad (1)
\displaystyle \text{It is given that } y(0)=1
\displaystyle \Rightarrow 1=0\cdot\sin^{-1}k+C
\displaystyle \Rightarrow C=1
\displaystyle \text{Substituting the value of } C \text{ in (1), we get}
\displaystyle y=x\sin^{-1}k+1
\displaystyle \Rightarrow y-1=x\sin^{-1}k
\displaystyle \text{Hence, } y-1=x\sin^{-1}k \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 23: }~e^{\,dy/dx}=x+1;\ y(0)=3.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle e^{\frac{dy}{dx}}=x+1
\displaystyle \text{Taking log on both sides, we get}
\displaystyle \frac{dy}{dx}\log e=\log(x+1)
\displaystyle \Rightarrow \frac{dy}{dx}=\log(x+1)
\displaystyle \Rightarrow dy=\{\log(x+1)\}\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\{\log(x+1)\}\,dx
\displaystyle \Rightarrow y=\int 1\cdot\log(x+1)\,dx
\displaystyle \Rightarrow y=\log(x+1)\int 1\,dx-\int\left[\frac{d}{dx}\{\log(x+1)\}\int 1\,dx\right]dx
\displaystyle \Rightarrow y=x\log(x+1)-\int\frac{x}{x+1}\,dx
\displaystyle \Rightarrow y=x\log(x+1)-\int\left(1-\frac{1}{x+1}\right)dx
\displaystyle \Rightarrow y=x\log(x+1)-x+\log(x+1)+C \qquad (1)
\displaystyle \text{It is given that } y(0)=3
\displaystyle \Rightarrow 3=0\cdot\log(0+1)-0+\log(0+1)+C
\displaystyle \Rightarrow C=3
\displaystyle \text{Substituting the value of } C \text{ in (1), we get}
\displaystyle y=x\log(x+1)+\log(x+1)-x+3
\displaystyle \Rightarrow y=(x+1)\log(x+1)-x+3
\displaystyle \text{Hence, } y=(x+1)\log(x+1)-x+3 \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 24: }~C'(x)=2+0.15x;\ C(0)=100.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle C'(x)=2+0.15x
\displaystyle \Rightarrow \frac{dC}{dx}=2+0.15x
\displaystyle \Rightarrow dC=(2+0.15x)\,dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dC=\int(2+0.15x)\,dx
\displaystyle \Rightarrow C=2x+\frac{0.15}{2}x^{2}+D \qquad (1)
\displaystyle \text{It is given that } C(0)=100
\displaystyle \Rightarrow 100=2(0)+\frac{0.15}{2}(0)^{2}+D
\displaystyle \Rightarrow D=100
\displaystyle \text{Substituting the value of } D \text{ in (1), we get}
\displaystyle C=2x+\frac{0.15}{2}x^{2}+100
\displaystyle \text{Hence, } C=2x+\frac{0.15}{2}x^{2}+100 \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 25: }~x\frac{dy}{dx}+1=0;\ y(-1)=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x\frac{dy}{dx}+1=0
\displaystyle \Rightarrow \frac{dy}{dx}=-\frac{1}{x}
\displaystyle \Rightarrow dy=\left(-\frac{1}{x}\right)dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left(-\frac{1}{x}\right)dx
\displaystyle \Rightarrow y=-\log|x|+C \qquad (1)
\displaystyle \text{It is given that } y(-1)=0
\displaystyle \Rightarrow 0=-\log|-1|+C
\displaystyle \Rightarrow C=0
\displaystyle \text{Substituting the value of } C \text{ in (1), we get}
\displaystyle y=-\log|x|
\displaystyle \text{Hence, } y=-\log|x| \text{ is the solution to the given differential equation.}

\displaystyle \textbf{Question 26: }~x(x^{2}-1)\frac{dy}{dx}=1;\ y(2)=0.
\displaystyle \text{Answer:}
\displaystyle \text{We have,}
\displaystyle x(x^{2}-1)\frac{dy}{dx}=1
\displaystyle \Rightarrow \frac{dy}{dx}=\frac{1}{x(x^{2}-1)}
\displaystyle \Rightarrow dy=\left\{\frac{1}{x(x^{2}-1)}\right\}dx
\displaystyle \text{Integrating both sides, we get}
\displaystyle \Rightarrow \int dy=\int\left\{\frac{1}{x(x^{2}-1)}\right\}dx
\displaystyle \Rightarrow y=\int\left\{\frac{1}{x(x^{2}-1)}\right\}dx
\displaystyle \Rightarrow y=\int\frac{1}{x(x-1)(x+1)}\,dx
\displaystyle \text{Let } \frac{1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}
\displaystyle \Rightarrow 1=A(x^{2}-1)+B(x^{2}+x)+C(x^{2}-x)
\displaystyle \Rightarrow 1=(A+B+C)x^{2}+(B-C)x-A
\displaystyle \text{Equating the coefficients on both sides, we get}
\displaystyle A+B+C=0 \qquad (1)
\displaystyle B-C=0 \qquad (2)
\displaystyle -A=1 \qquad (3)
\displaystyle \text{Solving (1), (2) and (3), we get}
\displaystyle A=-1,\ B=\frac{1}{2},\ C=\frac{1}{2}
\displaystyle \Rightarrow y=\int\left(\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\right)dx
\displaystyle \Rightarrow y=\int\left(-\frac{1}{x}+\frac{1}{2}\cdot\frac{1}{x-1}+\frac{1}{2}\cdot\frac{1}{x+1}\right)dx
\displaystyle \Rightarrow y=-\log|x|+\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|+C
\displaystyle \Rightarrow y=\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|-\log|x|+C
\displaystyle \text{It is given that } y(2)=0
\displaystyle \Rightarrow 0=\frac{1}{2}\log|2-1|+\frac{1}{2}\log|2+1|-\log|2|+C
\displaystyle \Rightarrow 0=\frac{1}{2}\log 1+\frac{1}{2}\log 3-\log 2+C
\displaystyle \Rightarrow C=\log 2-\frac{1}{2}\log 3
\displaystyle \text{Substituting the value of } C \text{, we get}
\displaystyle y=\frac{1}{2}\log|x-1|+\frac{1}{2}\log|x+1|-\log|x|+\log 2-\frac{1}{2}\log 3
\displaystyle \Rightarrow 2y=\log|x-1|+\log|x+1|-2\log|x|+2\log 2-\log 3
\displaystyle \Rightarrow 2y=\log\left(\frac{|x-1||x+1|}{|x|^{2}}\cdot\frac{4}{3}\right)
\displaystyle \Rightarrow 2y=\log\left(\frac{4(x^{2}-1)}{3x^{2}}\right)
\displaystyle \Rightarrow y=\frac{1}{2}\log\left(\frac{4(x^{2}-1)}{3x^{2}}\right)
\displaystyle \text{Hence, } y=\frac{1}{2}\log\left(\frac{4(x^{2}-1)}{3x^{2}}\right) \text{ is the solution to the given differential equation.}

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