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    solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))
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    (Original post by zomgleh)
    solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))
    Post your working so that appropriate hints can be given. If you can't see how to start, the obvious first step would be to divide by xy.
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    (Original post by zomgleh)
    solve the differential equation: xy dy/dx - y^2 = (3x^2)(e^(2y/x))
    Take out y as factor at LHS and divide by x^2
    \displaystyle \frac{y}{x^2}\cdot \left (x\frac{dy}{dx}-y\right )=3e^{2\frac{y}{x}}
    With substitution of
    \displaystyle t=\frac{y}{x}
    so
    \displaystyle y=t\cdot x
    and
    \displaystyle \frac{dt}{dx}=\frac{x\frac{dy}{d  x}-y}{x^2}
    The equation will be separable
    \displaystyle t\cdot x \frac{dt}{dx}=3e^{2t}
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    (Original post by Farhan.Hanif93)
    Post your working so that appropriate hints can be given. If you can't see how to start, the obvious first step would be to divide by xy.
    Okay, so I get dy/dx = 3xe^(2y/x) + (y/x)

    Am I right to say this equation is homogeneous of degree 0? The right hand side works out fine but I'm not sure about the dy/dx?
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    (Original post by zomgleh)
    Okay, so I get dy/dx = 3xe^(2y/x) + (y/x). Am I right to say this equation is homogeneous of degree 0?
    It will be if you manage to work out the RHS correctly. :P

    The right hand side works out fine but I'm not sure about the dy/dx?
    I'm not sure what you mean by "not sure about the dy/dx". A first order ODE is said to be homogeneous if dy/dx can be expressed as some function of (y/x); which, currently, it cannot (but will be possible if you fix the aforementioned error).
 
 
 
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