Question

For each of the differential equations, find the general solution:
1. $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$
2. $\frac{dy}{dx}=\sqrt{4-y^2}(-2<y<2)$
3. $\frac{dy}{dx}=1(y\ne 1)$
4. ${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$
5. $(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$
6. $\frac{dy}{dx}=(1+x^2)(1+y^2)$
7. $y\log ydx-xdy=0$
8. $x^4\frac{dy}{dx}=-y^5$
9. $\frac{dy}{dx}={\sin }^{-1}x$
10. $e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

Answer 1

1) $\frac{dy}{dx}=\frac{1-\cos x}{1+\cos x}$

$\int dy=\int \left[\frac{(1-\cos x)}{(1+\cos x)}\right)]dx$

$\int dy=\int \frac{2{\sin }^2\left(\frac{x}{2}\right)}{2{\cos }^2\left(\frac{x}{2}\right)}dx$

$\int dy=\int {\tan }^2\left(\frac{x}{2}\right)dx=\int \left({\sec }^2\right(\frac{x}{2}-1)dx$

$=\int {\sec }^2(\frac{x}{2}-1)dx$

$=2\tan (\frac{x}{2}-1)+c$

$y=2\tan (\frac{x}{2}-1)+c$

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

$\Rightarrow \frac{dy}{\sqrt{4-y^2}}=dx$

$\Rightarrow \frac{dy}{\sqrt{1-(\frac{y}{2}{)}^2}}=dx$

Let $\frac{y}{2}=t$

Thus $dy=2dt$

$\Rightarrow \frac{2dt}{\sqrt{1-t^2}}$

$\Rightarrow \int \frac{2dt}{\sqrt{1-t^2}}=\int dx$

$=2{\sin }^{-1}\left(f\right)=x+c$

$=2\sin y(\frac{y}{2}=x+c$

3) $\frac{dy}{dx}=1(y\ne 1)$

$\int dy=\int dx$

$y=x\ne c$

4) ${\sec }^{2}x\tan ydx+{\sec }^{2}y\tan xdy=0$

$\frac{dy}{dx}=\frac{{\sec }^{2}x\tan y}{{\sec }^{2}y\tan x}$

$\frac{{\sec }^{2}y}{\tan t}dy=-\frac{{\sec }^{2}x}{\tan x}dx$

$\tan y=u,\tan x=u$

${\sec }^{2}ydy=du$, ${\sec }^{2}xdx=dQ$

$\Rightarrow \int \frac{du}{u}=\frac{du}{u}$

$\log \left(u\right)=n-\log \left(Q\right)+\log C$

$\log \left(u\right)=\log \left(\frac{c}{Q}\right)$

$\log \left(uQ\right)=\log \left(C\right)$

$uQ=C$

$\tan y\tan x=C$

5) $(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

$(e^x+e^{-x})dy=(e^x-e^{-x})dx$

$\frac{dy}{dx}=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

$dy=\frac{e^x-e^{-x}}{e^x+e^{-x}}dx$

Integrating both sides.

$\int dx=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}dx$ ...(1)

$y=\int \frac{e^x-e^{-x}}{e^x+e^{-x}}$

Let $t=e^x+e^{-x}$

$\frac{dt}{dx}=(e^x-e^{-x})dx$

$dx=\frac{dt}{e^x-e^{-x}}$

Substituting values in (1), we get

$\int dy=\int \frac{e^x-e^{-x}}{t}\frac{dt}{e^x-e^{-x}}$

$\int dy=\int \frac{dt}{t}$

$y=\log |t|+C$

Putting back $t=e^x-e^{-x}$

$y=\log |e^x-e^{-x}|+C$

$y=\log (e^x-e^{-x})+C$ $(\therefore e^x-e^{-x}>0)$

6) $\frac{dy}{dx}=(1+x^2)(1+y^2)\Rightarrow dy$

$\Rightarrow \int \frac{1}{1+y^2}dy=\int (1+x^2)dx$

$\Rightarrow {\tan }^{-1}\left(y\right)=x+\frac{x^3}{3}+C$

7) $y\log ydx-xdy=0$

$y\log ydx=xdy$

$\frac{dx}{x}=\frac{dy}{y\log y}$

Intergrating both sides

$\int \frac{dy}{y\log y}=\int \frac{dx}{x}$

Put $t=\log y$

$dt=\frac{1}{y}dy$

$dy=ydt$

Hence, our equation becomes

$\int \frac{ydt}{y.t}=\int \frac{dx}{x}$

$\int \frac{dt}{t}=\int \frac{dx}{x}$

$\log |t|=\log |x|+\log c$

Putting $t=\log y$

$\log \left(\log y\right)=\log x+\log c$

$\log \left(\log y\right)=\log cx$ (Using $\log ab=\log a+\log b$)

$\log y=cx$

$y=e^{cx}$

8) $x^2\frac{dy}{dx}=-y^5$

$\int \frac{dy}{dx}=\int \frac{dx}{xy}\Rightarrow \int y^{-5}dy=-\int dx$

$\Rightarrow \frac{y^{-4}}{-4}=\frac{x^{-3}}{-3}+c$

$\Rightarrow \frac{1}{3x^3}+\frac{1}{4y^4}+C=0$

9) $\frac{dy}{dx}={\sin }^{-1}\left(x\right)\Rightarrow \int dy=\int {\sin }^{-1}\left(x\right)dx$

So $\therefore \int {\sin }^{-1}\left(x\right)dx=x{\sin }^{-1}\left(x\right)+\sqrt{1-x^2}$

$y=x{\sin }^{-1}\left(x\right)+\sqrt{1-x^2}+c$

10) $e^x\tan ydx+(1-e^x){\sec }^{2}ydy=0$

$e^x\tan ydx=(e^x-1){\sec }^{2}ydy$

$\int \frac{{\sec }^{2}y}{\tan y}dy=\int \frac{e^x}{e^x-1}dx$

$\tan y=u,e^x-1=Q$

${\sec }^{2}ydy=du$ $e^{x}dx=du$

$\int \frac{du}{u}=\int \frac{du}{u}$

$\log \left(u\right)=\log \left(Q\right)+\log \left(C\right)$

$\tan y=(e^x-1)=c$