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Math Help - Help with 1st order ODE of degree 4

  1. #1
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    Help with 1st order ODE of degree 4

    Hello everyone,

    I have this ODE and I am afraid I've got the wrong approach...
    Here's the DE:
    y'=\frac{y^{4}+x^{2}\,y^{2}-9\,x^{4}} {x\,y^{3}}

    I'm thinking that we may replace y' by dy/dx so that the DE becomes separable. Then we set x=vy so that we get rid of x and the only 2 variable remaining would be v and y...

    but I got to the point where -dy/dy=\frac{dv}{v^{3}\,(9v-1)}

    Have I got the right approach...
    Thanks for any help...
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  2. #2
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    Quote Originally Posted by rebghb View Post
    Hello everyone,

    I have this ODE and I am afraid I've got the wrong approach...
    Here's the DE:
    y'=\frac{y^{4}+x^{2}\,y^{2}-9\,x^{4}} {x\,y^{3}}

    I'm thinking that we may replace y' by dy/dx so that the DE becomes separable. Then we set x=vy so that we get rid of x and the only 2 variable remaining would be v and y...

    but I got to the point where -dy/dy=\frac{dv}{v^{3}\,(9v-1)}

    Have I got the right approach...
    Thanks for any help...
    Divide the numerator and denominator by y^4 and you get \frac{dy}{dx} = \frac{1 + \left( \frac{x}{y}\right)^2 - 9 \left( \frac{x}{y}\right)^4}{ \left( \frac{x}{y}\right)}.

    The usual technique is to now substitute \frac{y}{x} = v \Rightarrow y = x v \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}.

    Therefore v + x \frac{dv}{dx} = \frac{v^4 + v^2 - 9}{v^3}, which is seperable.
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