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Question

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


Solution

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

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

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

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

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

$$=2\tan \left (\dfrac{x}{2}-1\right)+c$$

$$y=2\tan \left (\dfrac{x}{2}-1\right)+c$$

 

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

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

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

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

Thus $$dy=2dt$$

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

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

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

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

 

3) $$\dfrac{dy}{dx} = 1  (y \neq 1)$$

$$\int dy = \int dx$$

$$y=x \neq c$$

 

4) $$\sec^2x \tan y dx+\sec^2y \tan x dy = 0$$

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

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

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

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

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

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

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

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

$$uQ = C$$

$$\tan y \tan x = C$$

 

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

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

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

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

Integrating both sides.

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

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

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

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

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

Substituting values in (1), we get

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

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

$$y=\log |t|+C$$

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

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

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

 

7) $$y \log y \,dx-x \,dy = 0$$

$$y \log y \,dx = x\, dy$$

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

Intergrating both sides

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

Put $$t =\log y$$

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

$$dy=y\,dt$$

Hence, our equation becomes

$$\int \dfrac{y\,dt}{y.t} = \int \dfrac{dx}{x}$$

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

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

Putting $$t=\log y$$

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

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

$$\log y=cx$$

$$y=e^{cx}$$


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

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

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

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


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

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

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

 

10) $$e^x\tan y \,dx +(1-e^x) \sec^2y dy = 0$$

$$e^x\tan y dx = (e^x-1)\sec^2y dy$$

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

$$\tan y = u,    e^x-1 = Q$$

$$\sec^2y\, dy = du$$  $$e^xdx=du$$

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

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

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


Mathematics

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