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For a given differential equation $$xy'+2y\log y -4x^2y = 0;$$ $$ y(1)=1 $$

I want to use the substitution $v=\log y$. Which implies that $$v'=\frac{dv}{dy}\frac{dy}{dx}= \frac{1}{y} y'$$Hence: $$y'=yv'$$

That being said, solving the equation I try as follows:

$$xyv'+2yv-4x^2y = 0 $$ $$xv'+2v = 4x^2 $$ $$v'+\frac{2}{x}v= 4x $$ $$v'=4x-\frac{2}{x}v $$

How can I use seperation of variables here? Or how can I solve this thing ?

thanks

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It is a liner equation, you do not separate the variables. –  Artem Jan 26 '13 at 13:33

1 Answer 1

up vote 1 down vote accepted

You've reduced your equation to the so called Bernoulli differential equation. Using notation from wikipedia just set $$ P(x)=\frac{2}{x}\qquad Q(x)=4x\qquad n=0 $$

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ah thanks. I remember that ;-) –  MSKfdaswplwq Jan 26 '13 at 13:37
    
You are weell come –  Norbert Jan 26 '13 at 13:39
    
@Norbert Why not Riccati equation with $q_2(x)=0$? –  Artem Jan 26 '13 at 13:45
    
@Artem Because Riccati equation is not allways solvable –  Norbert Jan 26 '13 at 14:09

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