Question

For each of the following differential equations, find the general solution:
(i) $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

(ii) $\frac{dy}{dx}=\sqrt{4-y^2},-2<y<2$

(iii) $\frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right)$

(iv) y log y dx − x dy = 0

(v) $\frac{dy}{dx}={\sin }^{-1}x$

(vi) $\frac{dy}{dx}+y=1$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\frac{1-\cos x}{1+\cos x} \\ \Rightarrow \frac{dy}{dx}=\frac{2{\sin }^2{\displaystyle \frac{x}{2}}}{2{\cos }^2{\displaystyle \frac{x}{2}}} \\ \Rightarrow \frac{dy}{dx}={\tan }^2\frac{x}{2} \\ \Rightarrow dy=\left({\tan }^2\frac{x}{2}\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left({\tan }^2\frac{x}{2}\right)dx \\ \Rightarrow \int dy=\int \left({\sec }^2\frac{x}{2}-1\right)dx \\ \Rightarrow y=2\tan \frac{x}{2}-x+C \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\sqrt{4-y^2} \\ \Rightarrow \frac{1}{\sqrt{4-y^2}}dy=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\sqrt{4-y^2}}dy=\int dx \\ \Rightarrow {\sin }^{-1}\frac{y}{2}=x+C \\ \Rightarrow \frac{y}{2}=\sin \left(x+C\right) \\ \Rightarrow y=2\sin \left(x+C\right) \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}=\left(1+x^2\right)\left(1+y^2\right) \\ \Rightarrow \frac{1}{\left(1+y^2\right)}dy=\left(1+x^2\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\left(1+y^2\right)}dy=\int \left(1+x^2\right)dx \\ \Rightarrow {\tan }^{-1}y=x+\frac{x^3}{3}+C \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right)\mathrm{We}\mathrm{have}, \\ y\log ydx-xdy=0 \\ \Rightarrow y\log ydx=xdy \\ \Rightarrow \frac{1}{x}dx=\frac{1}{y\log y}dy \\ \Rightarrow \frac{1}{y\log y}dy=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{y\log y}dy=\int \frac{1}{x}dx.....\left(1\right) \\ \mathrm{Putting}\log y=t \\ \Rightarrow \frac{1}{y}dy=dt \\ \mathrm{Therefore}\left(1\right)\mathrm{becomes} \\ \int \frac{1}{t}dt=\int \frac{1}{x}dx \\ \Rightarrow \log \left(t\right)=\log x+\log C \\ \Rightarrow \log \left(\log y\right)=\log x+\log C \\ \Rightarrow \log \left(\log y\right)=\log Cx \\ \Rightarrow \log y=Cx \\ \Rightarrow y=e^{Cx} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}={\sin }^{-1}x \\ \Rightarrow dy=\left({\sin }^{-1}x\right)dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int dy=\int \left({\sin }^{-1}x\right)dx \\ \Rightarrow \int dy=\int \underset{\mathrm{II}}{1}\times \underset{\mathrm{I}}{{\sin }^{-1}x}dx \\ \Rightarrow \int dy={\sin }^{-1}x\int 1dx-\int \left[\frac{d}{dx}\left({\sin }^{-1}x\right)\int 1dx\right]dx \\ \Rightarrow y=x{\sin }^{-1}x-\int \frac{x}{\sqrt{1-x^2}}dx \\ \mathrm{Putting}{t}^2=1-x^2,\mathrm{we}\mathrm{get} \\ 2tdt=-2xdx \\ \Rightarrow -tdt=xdx \\ \therefore y=x{\sin }^{-1}x+\int dt \\ \Rightarrow y=x{\sin }^{-1}x+t+C \\ \Rightarrow y=x{\sin }^{-1}x+\sqrt{1-x^2}+C \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right)\mathrm{We}\mathrm{have}, \\ \frac{dy}{dx}+y=1 \\ \Rightarrow \frac{dy}{dx}=1-y \\ \Rightarrow \frac{1}{\left(1-y\right)}dy=dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -\int \frac{1}{\left(y-1\right)}dy=\int dx \\ \Rightarrow \int \frac{1}{\left(y-1\right)}dy=-\int dx \\ \Rightarrow \log \left|y-1\right|=-x+\log C \\ \Rightarrow \log \left|y-1\right|-\log C=-x \\ \Rightarrow \log \left|\frac{y-1}{C}\right|=-x \\ \Rightarrow \frac{y-1}{C}=e^{-x} \\ \Rightarrow y=1+C{e}^{-x} \end{gathered}$$