I have the following differential equation (which is a simple one...but I am new to it). $$(x^2+2xy)y'=2(xy+y^2)$$ where $y(1)=2$ $$y'=\dfrac{2xy+2y^2}{2xy+x^2}=\dfrac{2 \left(\dfrac{y}{x}+\dfrac{y^2}{x^2}\right)}{2\dfrac{y}{x}+1}=\dfrac{2u+2u^2}{2u+1}$$ $$y=ux$$ $$u+xu'=\dfrac{2u+2u^2}{2u+1}$$ $$u'=\dfrac{1}{x}\left(\dfrac{2u+2u^2}{2u+1}-u\right)=\dfrac{1}{x}\dfrac{u}{2u+1}$$ Finally we get, separating variables $$2u+\ln |u| = \ln c|x|$$, where $c>0$ How am I supposed to find $u(x)$ from this equation? Finding $u$ would have my problem solved. Thank you!
Trouble point in differential equations
2
$\begingroup$
ordinary-differential-equations
derivatives
logarithms
absolute-value
homogeneous-equation
2 Answers
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$$2u+\ln |u| = \ln |c_1x|=\ln(x)+c_2$$ $$ue^{2u}=c_3x$$ $$2ue^{2u}=cx$$ $Xe^X=Y\quad\to\quad X=W(Y)\quad$ where $W$ is the Lambert W function :
http://mathworld.wolfram.com/LambertW-Function.html
With $X=2u\:$ and $\:Y=c\:x$ $$2u=W(cx)\quad\to\quad u=\frac{1}{2}W(c\:x)$$ $$y=\frac{x}{2}W(c\:x)$$
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From the last step : $2u+\ln |u| = \ln c|x|$
$e^{2u+\ln u}=e^{\ln cx}$
$ue^{2u}=cx.$ now put $u=\frac yx$
$ye^{\frac{2y}{x}}=cx^2$ is the implicit solution to your ODE.