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Question Number 57414 by Abdo msup. last updated on 03/Apr/19

solve y′ =2y^2 +y and y(o)=1

$${solve}\:\:{y}'\:=\mathrm{2}{y}^{\mathrm{2}} \:+{y}\:\:\:{and}\:{y}\left({o}\right)=\mathrm{1} \\ $$

Commented by kaivan.ahmadi last updated on 03/Apr/19

first we notice the integral ∫(dx/(2x^2 +x))=∫(dx/(x(2x+1)))= (A/x)+(B/(2x+1))=(((2A+B)x+A)/(x(2x+1)))=(1/(x(2x+1)))⇒ { ((2A+B=0)),((A=1)) :}⇒A=1,B=−2 ∫(dx/x)−∫((2dx)/(2x+1))=lnx−ln(2x+1)=ln((x/(2x+1))) now we solve the problem (dy/dx)=2y^2 +y⇒dx=(dy/(2y^2 +y))⇒ ∫dx=∫(dy/(2y^2 +y))⇒x=ln((y/(2y+1)))⇒ (y/(2y+1))=e^x ⇒2ye^x +e^x =y⇒ y(2e^x −1)=−e^x ⇒y=(e^x /(1−2e^x ))+C on the other hand y(0)=1 so 1=−1+C⇒C=2⇒ y=(e^x /(1−2e^x ))+2

$${first}\:{we}\:{notice}\:{the}\:{integral} \\ $$$$\int\frac{{dx}}{\mathrm{2}{x}^{\mathrm{2}} +{x}}=\int\frac{{dx}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}= \\ $$$$\frac{{A}}{{x}}+\frac{{B}}{\mathrm{2}{x}+\mathrm{1}}=\frac{\left(\mathrm{2}{A}+{B}\right){x}+{A}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}=\frac{\mathrm{1}}{{x}\left(\mathrm{2}{x}+\mathrm{1}\right)}\Rightarrow \\ $$$$\begin{cases}{\mathrm{2}{A}+{B}=\mathrm{0}}\\{{A}=\mathrm{1}}\end{cases}\Rightarrow{A}=\mathrm{1},{B}=−\mathrm{2} \\ $$$$\int\frac{{dx}}{{x}}−\int\frac{\mathrm{2}{dx}}{\mathrm{2}{x}+\mathrm{1}}={lnx}−{ln}\left(\mathrm{2}{x}+\mathrm{1}\right)={ln}\left(\frac{{x}}{\mathrm{2}{x}+\mathrm{1}}\right) \\ $$$${now}\:{we}\:{solve}\:{the}\:{problem} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{2}{y}^{\mathrm{2}} +{y}\Rightarrow{dx}=\frac{{dy}}{\mathrm{2}{y}^{\mathrm{2}} +{y}}\Rightarrow \\ $$$$\int{dx}=\int\frac{{dy}}{\mathrm{2}{y}^{\mathrm{2}} +{y}}\Rightarrow{x}={ln}\left(\frac{{y}}{\mathrm{2}{y}+\mathrm{1}}\right)\Rightarrow \\ $$$$\frac{{y}}{\mathrm{2}{y}+\mathrm{1}}={e}^{{x}} \Rightarrow\mathrm{2}{ye}^{{x}} +{e}^{{x}} ={y}\Rightarrow \\ $$$${y}\left(\mathrm{2}{e}^{{x}} −\mathrm{1}\right)=−{e}^{{x}} \Rightarrow{y}=\frac{{e}^{{x}} }{\mathrm{1}−\mathrm{2}{e}^{{x}} }+{C} \\ $$$${on}\:{the}\:{other}\:{hand}\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:{so} \\ $$$$\mathrm{1}=−\mathrm{1}+{C}\Rightarrow{C}=\mathrm{2}\Rightarrow \\ $$$${y}=\frac{{e}^{{x}} }{\mathrm{1}−\mathrm{2}{e}^{{x}} }+\mathrm{2} \\ $$

Commented by maxmathsup by imad last updated on 04/Apr/19

let use the changement y =(1/z) ⇒y^′ =−(z^′ /z^2 ) and (ed) ⇒−(z^′ /z^2 ) =(2/z^2 ) +(1/z) ⇒ −z^′ =2 +z ⇒z^′ +z +2 =0 y(0)=1 ⇒(1/(z(0))) =1 ⇒z(0)=1 (he) ⇒z^′ =−z ⇒−(z^′ /z) =1 ⇒−ln∣z∣=x+c_0 ⇒z(x) =k e^(−x) mvc method give z^′ =k^′ e^(−x) −k e^(−x) =(k^′ −k)e^(−x) and (e) ⇒ (k^′ −k)e^(−x) +ke^(−x) +2 =0 ⇒k^′ e^(−x) =−2 ⇒k^′ =−2 e^x ⇒ k =−2e^x +λ ⇒z(x) =(−2e^x +λ)e^(−x) =−2 +λ e^(−x) ⇒ y(x)=(1/(z(x))) =(1/(λe^(−x) −2)) y(o)=1 ⇒(1/(λ−2)) =1 ⇒λ−2 =1 ⇒λ =3 ⇒y(x)=(1/(3e^(−x) −2)) .

$${let}\:{use}\:{the}\:{changement}\:{y}\:=\frac{\mathrm{1}}{{z}}\:\Rightarrow{y}^{'} \:=−\frac{{z}^{'} }{{z}^{\mathrm{2}} }\:\:\:{and}\:\left({ed}\right)\:\Rightarrow−\frac{{z}^{'} }{{z}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{{z}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{{z}}\:\Rightarrow \\ $$$$−{z}^{'} \:=\mathrm{2}\:+{z}\:\Rightarrow{z}^{'} \:+{z}\:+\mathrm{2}\:=\mathrm{0}\:\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{{z}\left(\mathrm{0}\right)}\:=\mathrm{1}\:\Rightarrow{z}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$$\left({he}\right)\:\Rightarrow{z}^{'} \:=−{z}\:\Rightarrow−\frac{{z}^{'} }{{z}}\:=\mathrm{1}\:\Rightarrow−{ln}\mid{z}\mid={x}+{c}_{\mathrm{0}} \:\Rightarrow{z}\left({x}\right)\:={k}\:{e}^{−{x}} \: \\ $$$${mvc}\:{method}\:{give}\:\:{z}^{'} \:={k}^{'} \:{e}^{−{x}} \:−{k}\:{e}^{−{x}} \:=\left({k}^{'} −{k}\right){e}^{−{x}} \:\:{and}\:\left({e}\right)\:\Rightarrow \\ $$$$\left({k}^{'} −{k}\right){e}^{−{x}} \:+{ke}^{−{x}} \:+\mathrm{2}\:=\mathrm{0}\:\Rightarrow{k}^{'} \:{e}^{−{x}} \:=−\mathrm{2}\:\Rightarrow{k}^{'} \:=−\mathrm{2}\:{e}^{{x}} \:\Rightarrow \\ $$$${k}\:=−\mathrm{2}{e}^{{x}} \:+\lambda\:\Rightarrow{z}\left({x}\right)\:=\left(−\mathrm{2}{e}^{{x}} \:+\lambda\right){e}^{−{x}} \:=−\mathrm{2}\:+\lambda\:{e}^{−{x}} \:\Rightarrow \\ $$$${y}\left({x}\right)=\frac{\mathrm{1}}{{z}\left({x}\right)}\:=\frac{\mathrm{1}}{\lambda{e}^{−{x}} −\mathrm{2}} \\ $$$${y}\left({o}\right)=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\lambda−\mathrm{2}}\:=\mathrm{1}\:\Rightarrow\lambda−\mathrm{2}\:=\mathrm{1}\:\Rightarrow\lambda\:=\mathrm{3}\:\Rightarrow{y}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}{e}^{−{x}} −\mathrm{2}}\:. \\ $$

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