Math Is Fun Forum

Old_Steve

Member

Registered: 2007-05-15

Posts: 17

Differential Equation Help

I would appreciate any help on the following:

x+e^ydy-dx=0

I believe it is homogenous but I can't seem to solve it.

Thanks!

Last edited by Old_Steve (2007-05-15 12:47:10)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential Equation Help

Do you mean this differential equation?

(If that’s the one, separate the variables and integrate.)

Last edited by JaneFairfax (2007-05-16 03:29:36)

Stanley_Marsh

Member

Registered: 2006-12-13

Posts: 345

Re: Differential Equation Help

Last edited by Stanley_Marsh (2007-05-16 05:23:03)

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Numbers are the essence of the Universe

Old_Steve

Member

Registered: 2007-05-15

Posts: 17

Re: Differential Equation Help

It is actually listed in the text as I have entered it.  An alternate form may be x + e^y =dx/dy.   The answer is x=ce^y+ye^y but I don't know how to get there.  I'm going in circles.

Thanks for any help!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential Equation Help

Ah, so it was:

Well,

And now we just solve this second-order differential equation.

It turns out to be

However the original differential equation is a first-order equation, so we must eliminate one of the arbitrary constants. Differentiate:

This must be equal to

for all y. Comparing the coefficients, B = 0.

Hence the solution to the differential equation is

Last edited by JaneFairfax (2007-05-16 08:35:20)

Old_Steve

Member

Registered: 2007-05-15

Posts: 17

Re: Differential Equation Help

Well...unfortunately you lost me.    We have not started second order differential equations yet and I don't think I am supposed to solve it that way.  I'll keep trying.  Thanks for your help!!!!

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential Equation Help

You’re welcome. I actually think second-order equations are much easier to solve than first-order ones. In particular, suppose you have

where a and b are constants. This is called a homogeneous second-order linear ordinary differential equation. To solve it, you consider the auxiliary equation m[sup]2[/sup] + am + b = 0. Suppose the roots are α and β. The general solution is

To solve the inhomogeneous second-order linear ordinary differential equation

(where F is a function of t) look for any function ψ satisfying the above equation, i.e.

ψ is called a particular solution; finding particular solutions is usually quite straightforward (as you’ll no doubt learn in due course). The general solution to the inhomogeous equation is then the general solution to the homogeous equation plus the particular solution.

Last edited by JaneFairfax (2007-05-16 23:17:35)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Differential Equation Help

It's the first time I've ever heard that the second order is easier!!

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X'(y-Xβ)=0

Old_Steve

Member

Registered: 2007-05-15

Posts: 17

Re: Differential Equation Help

I actually had to go back and work it as a first order equation.  It fits the form x' + p(y)x = g(x).  I would enter my work but I don't know where you all get your fomula editors.  Anyway, I was able to arrive at the correct answer by working it as a firt order equation.

Last edited by Old_Steve (2007-05-20 22:08:10)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential Equation Help

It fits the form x' + p(y)x = g(x).

You mean g(y).

Yeah, I just realized how to solve it as a first-order equation as well. You just solve the homogenous case x′ + p(y)x = 0 (which is a variables-separable equation). Then you find a particular solution to x′ + p(y)x = g(y). The general solution to x′ + p(y)x = g(y) is then solution to general case + particular solution.

Is that how you did it?

Old_Steve

Member

Registered: 2007-05-15

Posts: 17

Re: Differential Equation Help

Yes!  It should have been g(y).  I was able to solve it as a first order linear diff eq. with u(y)=e^-y.   Multiply through and integrate.  Then multiply trough again and the answer is x = e^y[y +c].

Last edited by Old_Steve (2007-05-21 13:08:39)

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Differential Equation Help

Hey, that works as well! Great job.

Last edited by JaneFairfax (2007-05-21 10:17:55)

George,Y

Member

Registered: 2006-03-12

Posts: 1,379

Re: Differential Equation Help

Yes, Linear!
-1(x)
So e[sup]-y[/sup]|'[sub]y[/sub]=e[sup]-y[/sup](-1)

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X'(y-Xβ)=0