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(Moderator's note: thread moved from "Differential Equations")

Hello :)

I'm trying to solve an homogeneous equation... but it seems that i'm wrong in some step... or something, because I can't complete this problem, look here is what i got:

[tex]x\frac{dy}{dx} = ye^\frac{x}{y} - x[/tex]

[tex]x dy = (ye^\frac{x}{y} - x) dx[/tex]

[tex]x dy + ( -ye^\frac{x}{y} + x) dx[/tex]

Using substitution

[tex]x = vy[/tex]

[tex]dx = vdy + ydv[/tex]

[tex]vy dy + (-ye^v + vy) [ vdy + ydv ] = 0[/tex]

[tex](vy + yv^2 - vye^v)dy + (vy^2 - y^2e^v )dv = 0[/tex]

[tex](vy + yv^2 - vye^v)dy = (y^2e^v - vy^2 )dv[/tex]

[tex]\frac{(y)(v + v^2 - ve^v)}{y^2(e^v - v)} dy = dv[/tex]

[tex]\int \frac{1}{y} dy = \int \frac{(e^v - v)}{(v + v^2 - ve^v)} dv[/tex]

The integral on the left side is easy to solve, but I can't find a way to solve the right side of the equation.

Any Suggestions?

Thanks and sorry for my English, I'm still learning it (i'm from venezuela)

Hello :)

I'm trying to solve an homogeneous equation... but it seems that i'm wrong in some step... or something, because I can't complete this problem, look here is what i got:

[tex]x\frac{dy}{dx} = ye^\frac{x}{y} - x[/tex]

[tex]x dy = (ye^\frac{x}{y} - x) dx[/tex]

[tex]x dy + ( -ye^\frac{x}{y} + x) dx[/tex]

Using substitution

[tex]x = vy[/tex]

[tex]dx = vdy + ydv[/tex]

[tex]vy dy + (-ye^v + vy) [ vdy + ydv ] = 0[/tex]

[tex](vy + yv^2 - vye^v)dy + (vy^2 - y^2e^v )dv = 0[/tex]

[tex](vy + yv^2 - vye^v)dy = (y^2e^v - vy^2 )dv[/tex]

[tex]\frac{(y)(v + v^2 - ve^v)}{y^2(e^v - v)} dy = dv[/tex]

[tex]\int \frac{1}{y} dy = \int \frac{(e^v - v)}{(v + v^2 - ve^v)} dv[/tex]

The integral on the left side is easy to solve, but I can't find a way to solve the right side of the equation.

Any Suggestions?

Thanks and sorry for my English, I'm still learning it (i'm from venezuela)

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