- #1

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[tex]\frac{du}{dx} = \frac{4u-2x}{u+x}[/tex]

Whenever I try to complete it I get a function which I don't think should be integratible. Could some light be shed onto this matter?

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- Thread starter drcameron
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- #1

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- 0

[tex]\frac{du}{dx} = \frac{4u-2x}{u+x}[/tex]

Whenever I try to complete it I get a function which I don't think should be integratible. Could some light be shed onto this matter?

- #2

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[tex]\frac{du}{dx}=\frac{4\frac{u}{x}-2}{\frac{u}{x}+1}[/tex]

This is a homogeneous differential equation, which means the right side can be expressed as a function of u/x. The way to deal with equations like that are to perform the next change of variables:

[tex] z=\frac{u}{x} => u=xz [/tex]

[tex] \frac{du}{dx}=x\frac{dz}{dx}+z [/tex]

Then, if you have a general homogeneous equation:

[tex] \frac{du}{dx}=F(\frac{u}{x})[/tex]

It turns into:

[tex] x\frac{dz}{dx}+z=F(z) [/tex]

And by seperation of variables method:

[tex] \frac{dz}{F(z)-z}=\frac{dx}{x} [/tex]

You integrate to retrieve z(x) (commonly you will have an implicit function form), and change back to u(x).

Notice that you'll probably have singular solutions for z, and that constant singular solution for z are linear singular solutions for u.

- #3

HallsofIvy

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If we let v= u/x, then u= xv and du/dx= x dv/dx+ v so the equation becomes

[tex]x\frac{dv}{dx}+ v= \frac{4v-2}{v+1}[/tex]

then

[tex]x\frac{dv}{dx}= \frac{4v-2}{v+1}- v= -\frac{v^2- 4v-2}{v+1}[/tex]

so, separating,

[tex]\frac{v+1}{v^2- 4v- 2} dv= \frac{1}{x}dx[/tex]

Now that certainly is integrable: use "partial fractions" on the left. The denominator factors as [itex](v- 2+\sqrt{6})(v-2-\sqrt{6})[/itex]. I got that by completing the square: [itex]v^2- 4v- 2= v^2- 4v+ 4- 4- 2= (v-2)^2- 6[/itex] which can be factored s a "difference of squares".

- #4

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For some reason I had convinced myself to use y=u+x as a substition - its easy to see where i went wrong then!

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