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Question Number 823 by 112358 last updated on 18/Mar/15

Solve the following d.e by using  v=(dy/dx), where v is a function of x.  x^2 (d^2 y/dx^2 )−2(dy/dx)+x=0

$${Solve}\:{the}\:{following}\:{d}.{e}\:{by}\:{using} \\ $$$${v}=\frac{{dy}}{{dx}},\:{where}\:{v}\:{is}\:{a}\:{function}\:{of}\:{x}. \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\frac{{dy}}{{dx}}+{x}=\mathrm{0} \\ $$

Answered by 123456 last updated on 18/Mar/15

v=(dy/dx)⇒(dv/dx)=(d^2 y/dx^2 )  x^2 (d^2 y/dx^2 )−2(dy/dx)+x=0⇒x^2 (dv/dx)−2v+x=0  by the coment  v=−e^(−(2/x)) ∫(e^(2/x) /x)dx  (dy/dx)=−e^(−(2/x)) ∫(e^(2/x) /x)dx  y=∫−e^(−(2/x)) ∫(e^(2/x) /x)dxdx  y=−∫e^(−(2/x)) ∫(e^(2/x) /x)dxdx  Ei(u)=∫(e^u /u)du  ∫(e^(2/x) /x)dx  u=(2/x)⇔x=(2/u)  du=−((2dx)/x^2 )⇔−((xdu)/2)=(dx/x)⇔−(du/u)=(dx/x)  −∫(e^u /u)du=−Ei(u)+C_1 =−Ei((2/x))+C_1   y=∫−e^(−(2/x)) [−Ei((2/x))+C_1 ]dx  y=∫e^(−(2/x)) Ei((2/x))dx−C_1 ∫e^(−(2/x)) dx

$${v}=\frac{{dy}}{{dx}}\Rightarrow\frac{{dv}}{{dx}}=\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\mathrm{2}\frac{{dy}}{{dx}}+{x}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} \frac{{dv}}{{dx}}−\mathrm{2}{v}+{x}=\mathrm{0} \\ $$$${by}\:{the}\:{coment} \\ $$$${v}=−{e}^{−\frac{\mathrm{2}}{{x}}} \int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx} \\ $$$$\frac{{dy}}{{dx}}=−{e}^{−\frac{\mathrm{2}}{{x}}} \int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx} \\ $$$${y}=\int−{e}^{−\frac{\mathrm{2}}{{x}}} \int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dxdx} \\ $$$${y}=−\int{e}^{−\frac{\mathrm{2}}{{x}}} \int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dxdx} \\ $$$$\mathrm{E}{i}\left({u}\right)=\int\frac{{e}^{{u}} }{{u}}{du} \\ $$$$\int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx} \\ $$$${u}=\frac{\mathrm{2}}{{x}}\Leftrightarrow{x}=\frac{\mathrm{2}}{{u}} \\ $$$${du}=−\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} }\Leftrightarrow−\frac{{xdu}}{\mathrm{2}}=\frac{{dx}}{{x}}\Leftrightarrow−\frac{{du}}{{u}}=\frac{{dx}}{{x}} \\ $$$$−\int\frac{{e}^{{u}} }{{u}}{du}=−\mathrm{Ei}\left(\mathrm{u}\right)+\mathrm{C}_{\mathrm{1}} =−\mathrm{Ei}\left(\frac{\mathrm{2}}{{x}}\right)+\mathrm{C}_{\mathrm{1}} \\ $$$${y}=\int−{e}^{−\frac{\mathrm{2}}{{x}}} \left[−\mathrm{Ei}\left(\frac{\mathrm{2}}{{x}}\right)+\mathrm{C}_{\mathrm{1}} \right]{dx} \\ $$$${y}=\int{e}^{−\frac{\mathrm{2}}{{x}}} \mathrm{Ei}\left(\frac{\mathrm{2}}{{x}}\right){dx}−\mathrm{C}_{\mathrm{1}} \int{e}^{−\frac{\mathrm{2}}{{x}}} {dx} \\ $$

Commented by 123456 last updated on 18/Mar/15

x^2 (dy/dx)−2y+x=0  y=uv,v≠0  (dy/dx)=(du/dx)v+u(dv/dx)  x^2 ((du/dx)v+u(dv/dx))−2uv+x=0  v(x^2 (du/dx)−2u)+(x^2 u(dv/dx)+x)=0  seting the red part 0  x^2 (du/dx)−2u=0  x^2 du=2udx  (du/u)=((2dx)/x^2 )  ∫(du/u)=∫((2dx)/x^2 )=2∫(dx/x^2 )  ln u=−(2/x)+A_1   u=A_2 e^(−(2/x))   and then  x^2 u(dv/dx)+x=0  x^2 udv=−xdx  dv=−(dx/(ux))  ∫dv=∫−(dx/(ux))=−∫(dx/(ux))  v=−(1/A_2 )∫(e^(2/x) /x)dx  y=uv=(A_2 e^(−(2/x)) )×(−(1/A_2 )∫(e^(2/x) /x)dx)  y=−e^(−(2/x)) ∫(e^(2/x) /x)dx

$${x}^{\mathrm{2}} \frac{{dy}}{{dx}}−\mathrm{2}{y}+{x}=\mathrm{0} \\ $$$${y}={uv},{v}\neq\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}{v}+{u}\frac{{dv}}{{dx}} \\ $$$${x}^{\mathrm{2}} \left(\frac{{du}}{{dx}}{v}+{u}\frac{{dv}}{{dx}}\right)−\mathrm{2}{uv}+{x}=\mathrm{0} \\ $$$${v}\left({x}^{\mathrm{2}} \frac{{du}}{{dx}}−\mathrm{2}{u}\right)+\left({x}^{\mathrm{2}} {u}\frac{{dv}}{{dx}}+{x}\right)=\mathrm{0} \\ $$$$\mathrm{seting}\:\mathrm{the}\:\mathrm{red}\:\mathrm{part}\:\mathrm{0} \\ $$$${x}^{\mathrm{2}} \frac{{du}}{{dx}}−\mathrm{2}{u}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} {du}=\mathrm{2}{udx} \\ $$$$\frac{{du}}{{u}}=\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} } \\ $$$$\int\frac{{du}}{{u}}=\int\frac{\mathrm{2}{dx}}{{x}^{\mathrm{2}} }=\mathrm{2}\int\frac{{dx}}{{x}^{\mathrm{2}} } \\ $$$$\mathrm{ln}\:{u}=−\frac{\mathrm{2}}{{x}}+\mathrm{A}_{\mathrm{1}} \\ $$$${u}=\mathrm{A}_{\mathrm{2}} {e}^{−\frac{\mathrm{2}}{{x}}} \\ $$$$\mathrm{and}\:\mathrm{then} \\ $$$${x}^{\mathrm{2}} {u}\frac{{dv}}{{dx}}+{x}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} {udv}=−{xdx} \\ $$$${dv}=−\frac{{dx}}{{ux}} \\ $$$$\int{dv}=\int−\frac{{dx}}{{ux}}=−\int\frac{{dx}}{{ux}} \\ $$$${v}=−\frac{\mathrm{1}}{\mathrm{A}_{\mathrm{2}} }\int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx} \\ $$$${y}={uv}=\left(\mathrm{A}_{\mathrm{2}} {e}^{−\frac{\mathrm{2}}{{x}}} \right)×\left(−\frac{\mathrm{1}}{\mathrm{A}_{\mathrm{2}} }\int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx}\right) \\ $$$${y}=−{e}^{−\frac{\mathrm{2}}{{x}}} \int\frac{{e}^{\frac{\mathrm{2}}{{x}}} }{{x}}{dx} \\ $$

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