# Differential Equations?

find a differential equation that has y=c1exp(2t) +c2exp(-3t)

I take derviatives and try to elimminate stuff, but I can't make it go away........ I know the anser, but I can't seem to same thing..

Update:

y=c1exp(2t) +c2exp(-37) is the solution to some differential equation....I am looking for a differential equation that this solution would satisfy..

Relevance

Man, Ippus just beat me to it. :c( i have an easier solution at the bottom, though.

There are two roots, so you're looking for a 2nd order differential equation.

Differentiate once:

y' = 2 c1 e^(2t) - 3 c2 e^(-3t)

And then again:

y'' = 4 c1 e^(2t) + 9 c2 e^(-3t)

And then express y'' using y and y':

y'' = ay' + by

4 c1 e^(2t) + 9 c2 e^(-3t) = a[2 c1 e^(2t) - 3 c2 e^(-3t)] + b[c1 e^(2t) + c2 e^(-3t)]

To make things a little easier, break this into two equations:

4 c1 e^(2t) = a[2 c1 e^(2t) ] + b[c1 e^(2t)]

9 c2 e^(-3t) = a[-3 c2 e^(-3t)] + b[c2 e^(-3t)]

And then you can make things even easier by dividing the 1st equation by c1 e^(2t) and the second by c2 e^(-3t):

4 = 2a + b

9 = -3a + b

Subtract the 2nd equation from the first, and you get 5a = -5, so a = -1 and b = 4 - 2(-1) = 6.

So the differential equation would be:

y = -y' + 6y

*******************************************

That's the hard way.

The easy way is just to know that, 2 and -3 must be roots of the characteristic equation:

(r - 2)(r + 3) = r² + r - 6

And this corresponds to the differential equation:

y'' + y' - 6y = 0

y'' = -y' + 6y

The presence of two terms in the expression for y means you'll need to look for a second order differential equation.

Perhaps this is what was throwing you off?

It will be easier to read my post if I write your c1 coefficient as c, and c2 as d.

y = c e^2t + d e^-3t

differentiate w.r.t. t:

y' = 2c e^2t -3d e^-3t

and again:

y'' = 4c e^2t + 9d e^-3t

For this to be written as a linear function of y and y' (we expect this to be the case given the simple exponential form of the solution):

y'' = A y + B y'

for some coefficients A and B, where

4 = A + 2B (1)

9 = A - 3B (2)

This is a pair of simultaneous algebraic equations.

By elimination: (2) => A = 9+3B

so (1) gives us

4 = 9+ 3B + 2B

-5 = 5B

B = -1

which implies A = 6.

so y'' = 6y - y'

i.e., y'' + y' - 6y = 0

You can verify this by plugging your y expression back in to this differential equation and getting 0!

Source(s): I'm a mathematician

Post what you 'know' is the answer and I'll get back to you.