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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