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..
y=c1exp(2t) +c2exp(-37) is the solution to some differential equation....I am looking for a differential equation that this solution would satisfy..
- Jim BurnellLv 61 decade agoFavorite Answer
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.
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
- 1 decade ago
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
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
- modulo_functionLv 71 decade ago
Post what you 'know' is the answer and I'll get back to you.
- Anonymous1 decade ago
I am also not sure about it..
Will u plz tell me what is the answer?