Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

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)) .

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com