- #1

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u"+(u-1)u=0

I tried substuting u'=v but got stuck when I wasn't given initial values and couldn't solve for the constants after integrating.

- Thread starter 1weasel
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- #1

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u"+(u-1)u=0

I tried substuting u'=v but got stuck when I wasn't given initial values and couldn't solve for the constants after integrating.

- #2

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If you use the substitution:

u"+(u-1)u=0

I tried substuting u'=v but got stuck when I wasn't given initial values and couldn't solve for the constants after integrating.

[tex]u=y[/tex]

[tex]v=y'[/tex]

[tex]\frac{dv}{du}=\frac{y''}{y'}[/tex]

from which we have:

[tex]y=u[/tex]

[tex]y'=v[/tex]

[tex]y''=v\frac{dv}{du}[/tex]

You end up with a first order DE:

[tex]vdv=(1-u)udu[/tex]

Which is separable. The inverse substitution in the solution gives then a new first order equation in [itex]y'[/itex], which can be solved because it is also separable. However the integral is complicated. The end solution will be:

[tex]x=\int\frac{dy}{\sqrt{y^2-\frac{2}{3}y^3+K_1}}[/tex]

Perhaps you have boundary conditions which can make life easier, although I think it will stay an unpleasant integral. Hope this helps a bit.

Last edited:

- #3

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I rewrote the equation as:

[tex]\frac{d^2y}{dx^2}+y(y-1)=0[/tex]

- #4

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y" - y' -2y = F(t); where y(0) = 0, y'(0) = 0 and F(t) = 1 if 0<= t < 2

= t^2 + 1 if 2 <= t < 5

= t^2 + t if t => 5

I'm not having problem setting up but having trouble once I have everything on the right hand side. Unless I'm not using me brain the partial fractions look insane.

- #5

HallsofIvy

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