Inhomogeneous differential equation, boundary

1. Oct 18, 2009

elmarsur

y"+by'+c=0
y(0)=y(1)=0

2. Oct 19, 2009

gato_

Actually, the equation only implies y', so the equation for z=y' is first order and can be directly integrated by variation of constants (assume the same solution as for the homogeneous equation, z=C*exp(-b/a *x), and the assume C is a function of x). solve for C, and then you will get z=y'. integrate this and you will get y

3. Oct 19, 2009

HallsofIvy

Staff Emeritus
Do you have a question about this? It looks like homework to me and so belongs in the homework section. What have YOU done on it. It is a relatively simple linear equation with constant coefficients. What is the associated homogeneous equation? What is its characteristic equation?

4. Oct 19, 2009

elmarsur

Thanks a million Gato!
==========================================
No, Hall. It's not homework. Probably the terse manner in which I plastered on the board gave that impression.

5. Oct 19, 2009

elmarsur

This is what I had done previously:

Take the homogeneous side:y"+by'=0
Condition: y is never zero.
General solution for homogeneous: y = e^mx
Substituting: (m^2+bm)e^mx = 0
e^mx is never zero.
Then: m^2 + bm = 0 (characteristic equation)
b never zero (condition above).
Then: m(m+b)=0; m=0; m=-b
y=c1 + c2e^(-bx)
y(0)=0; then, c1 + c2 = 0
y(1)=0; then, c2e^(-b)= 0

I have a non-trivial solution only if the det: [1 1]
[0 e^(-b)]
{which is equal to e^(-b)} is zero.
If not, then y=0 (trivial solution)

But this is a Reynold's equation (fluid dynamics) and the function cannot be zero throughout

Last edited: Oct 19, 2009
6. Oct 20, 2009

HallsofIvy

Staff Emeritus
the theorem you state, that there is a non-trivial solution only if the Jacobian is zero is only for homogeneous equations. The trivial solution is always a solution to homogeneous equations so there is a non-trival solution only if the solution is not unique. That is not the case for non-homogeneous equations as you have here.

Okay, the characteristic equation is $r^2+ br= r(r+b)= 0$ and so the characteristic roots are 0 and -b. The general solution to the associated homogeneous equation is $y= A+ Be^{-bt}$.

Since a constant already satisfies the homogeneous equations, try a particular solution of the form Pt. Then y'= P, y"= 0 so y"+ y'= bP= c. P= c/b and the particular solution is (c/b)t.

The general solution to the entire equation, then, is y(t)= A+ Be^{-bt}+ (c/b)t.

Setting t= 0, y(0)= A= 0.

Setting t= 1, $y(1)= Be^{-b}+ (c/b)= 0$ so $Be^{-b}= -c/b$ and $B= (-c/b)e^{b}$.

$y(t)= (-c/b)e^{b}e^{-bt}+ (c/b)t= (c/b)(t- e^{b(1- t)})$ is the only solution to that problem.

Last edited: Oct 20, 2009
7. Oct 20, 2009

elmarsur

Thanks Hall. I had stopped at the homogeneous because I didn't think of the Pt type solution of the inhomogeneous.
However, for t=0, y(0)=0= A+B , instead of A (as you wrote). (1/(e^0)=1)
Right?
Or have I lost all my faculties?