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Question Number 96192 by 1549442205 last updated on 30/May/20
find a particular solution to the equation y′ =(y/x)+sin(y/x) with original condition y(1)=(π/2)
$$\mathrm{find}\:\mathrm{a}\:\mathrm{particular}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\mathrm{y}'\:=\frac{\mathrm{y}}{\mathrm{x}}+\mathrm{sin}\frac{\mathrm{y}}{\mathrm{x}}\:\mathrm{with}\:\mathrm{original}\:\mathrm{condition} \\ $$$$\mathrm{y}\left(\mathrm{1}\right)=\frac{\pi}{\mathrm{2}} \\ $$
Commented by john santu last updated on 30/May/20
set v =(y/x) ⇒(dy/dx) = v +x (dv/dx) v+x (dv/dx) = v +sin v x (dv/dx) = sin v ⇒ ∫ (dv/(sin v)) = ∫ (dx/x) ∫ csc v dv = ln(x) + c ln (csc v−cot v) = ln Cx csc v −cot v = Cx ((1−cos ((y/x)))/(sin ((y/x)))) = Cx 1−cos ((y/x)) = Cx sin ((y/x))
$$\mathrm{set}\:{v}\:=\frac{{y}}{{x}}\:\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:{v}\:+{x}\:\frac{{dv}}{{dx}}\: \\ $$$${v}+{x}\:\frac{{dv}}{{dx}}\:=\:{v}\:+\mathrm{sin}\:{v}\: \\ $$$${x}\:\frac{{dv}}{{dx}}\:=\:\mathrm{sin}\:{v}\:\Rightarrow\:\int\:\frac{{dv}}{\mathrm{sin}\:{v}}\:=\:\int\:\frac{{dx}}{{x}} \\ $$$$\int\:\mathrm{csc}\:{v}\:{dv}\:=\:\mathrm{ln}\left({x}\right)\:+\:{c} \\ $$$$\mathrm{ln}\:\left(\mathrm{csc}\:{v}−\mathrm{cot}\:{v}\right)\:=\:\mathrm{ln}\:\mathrm{C}{x} \\ $$$${csc}\:{v}\:−\mathrm{cot}\:{v}\:=\:\mathrm{C}{x} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\left(\frac{{y}}{{x}}\right)}{\mathrm{sin}\:\left(\frac{{y}}{{x}}\right)}\:=\:{Cx}\: \\ $$$$\mathrm{1}−\mathrm{cos}\:\left(\frac{{y}}{{x}}\right)\:=\:{Cx}\:\mathrm{sin}\:\left(\frac{{y}}{{x}}\right) \\ $$$$ \\ $$
Commented by 1549442205 last updated on 11/Jun/20
This is an other way Put (y/x)=t⇒y=tx⇒dy=xdt+tdx.Hence, xdt+tdx=(t+sint)dx⇒xdt=sintdx ⇒(dt/(sint))=(dx/x).Integrate two sides we get ln∣tan((t/2))∣=ln∣x∣+lnC.From that∣ (t/2)=arctan(Cx)⇒y=2x.arctan(Cx) Using the original condition we get (π/2)=2arctanC⇒C=1.Thus,the particular solution has form: y=2x.arctanx
$$\mathrm{This}\:\mathrm{is}\:\mathrm{an}\:\mathrm{other}\:\mathrm{way} \\ $$$$\mathrm{Put}\:\frac{\mathrm{y}}{\mathrm{x}}=\mathrm{t}\Rightarrow\mathrm{y}=\mathrm{tx}\Rightarrow\mathrm{dy}=\mathrm{xdt}+\mathrm{tdx}.\mathrm{Hence}, \\ $$$$\mathrm{xdt}+\mathrm{tdx}=\left(\mathrm{t}+\mathrm{sint}\right)\mathrm{dx}\Rightarrow\mathrm{xdt}=\mathrm{sintdx} \\ $$$$\Rightarrow\frac{\mathrm{dt}}{\mathrm{sint}}=\frac{\mathrm{dx}}{\mathrm{x}}.\mathrm{Integrate}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{we}\:\mathrm{get} \\ $$$$\mathrm{ln}\mid\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)\mid=\mathrm{ln}\mid\mathrm{x}\mid+\mathrm{lnC}.\mathrm{From}\:\mathrm{that}\mid \\ $$$$\frac{\mathrm{t}}{\mathrm{2}}=\mathrm{arctan}\left(\mathrm{Cx}\right)\Rightarrow\mathrm{y}=\mathrm{2x}.\mathrm{arctan}\left(\mathrm{Cx}\right) \\ $$$$\mathrm{Using}\:\mathrm{the}\:\mathrm{original}\:\mathrm{condition}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\pi}{\mathrm{2}}=\mathrm{2arctanC}\Rightarrow\mathrm{C}=\mathrm{1}.\mathrm{Thus},\mathrm{the}\: \\ $$$$\mathrm{particular}\:\mathrm{solution}\:\mathrm{has}\:\mathrm{form}: \\ $$$$\boldsymbol{\mathrm{y}}=\mathrm{2}\boldsymbol{\mathrm{x}}.\boldsymbol{\mathrm{arctanx}} \\ $$

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