Help solving 3rd order differential equation?

Question

I need help solving the following diffEQ:


y'''+2y''-5y'-6y=100e^-3x+18^-x

no other information provided, ie. initial values.

I have only seen examples where the roots are all the same. In this example, its @ 2, -1, -3.

How do I work it from here?

yahoo cut off my question.

the diff EQ is:

y'''+2y''-5y'-6y=
100e^-3x+18e^-x

Had to go on two lines so it didnt mess up

Answer 1

Because you specified the roots, you clearly know the characteristic polynomial for this differential equation is:

r^3 + 2*r^2 - 5*r - 6

Remember how you got this. You got this by assuming that y=e^(r*x) and substituted this into your differential equation. Thus, your HOMOGENOUS solution will be a linear combination of e^(r*x) terms. We'll call this solution the CHARACTERISTIC (or complimentary) solution.

The roots of the characteristic equation are, as you say, 2, -1, and -3. That means your CHARACTERISTIC (or "complimentary") solution is:

yc(t) = c1*e^(2x) + c2*e^(-1x) + c3*e^(-3x)

where c1, c2, and c3 are all constants. (which you won't ever know without initial conditions, so leave them as constants) Notice that the coefficient in the argument of the exponentials is just the root multiplied by x (or t or whatever your independent variable is). THIS DOES NOT COMPLETE YOUR PROBLEM.

Note that this is the solution to the HOMOGENOUS version of this differential equation. That is, yc(x) is the solution to:

yc''' + 2yc'' - 5yc' - 6yc = 0

That's identical to your problem, except that the right-hand side is 0.

The general solution to your NONHOMOGENOUS differential equation is:

y(x) = yp(x) + yc(x)

where yp(x) is the PARTICULAR solution to the problem. You will either have to use the method of "undetermined coefficients" or the method of "variation of parameters" with this problem. the method of "undetermined coefficients" assumes that the solution takes a form similar to the right-hand side of the problem and solves for coefficients in that solution. Unfortunately, this will not work for this problem.

IF YOU TRY to use the method of UNDETERMIEND COEFFICIENTS, you will get the result that:

0*A = 100
0*B = 18

where A and B are your two coefficients. Clearly, A and B cannot be determined! DO NOT make the mistake in assuming that A=B=0! yp=0 is *NOT* a solution to this problem!! If your differential equation was a physical system, the expression on the right-hand side would be something driving the system. This driving force will not lead to a yp=0 answer.

Thus, you **need** to use VARIATION OF PARAMETERS to solve the problem.

See the FIRST source below for instructions on how to solve this problem using the variation of parameters. You will need to use your CHARACTERISTIC solution yc(x) above to use this method.

Once you have both yp(x) and yc(x), you'll be done and your solution will be:

y(x) = yp(x) + yc(x)

The other two sources describe how to use the other method (undetermined coefficients) and some other general information about differential equations.

Answer 2

For the inhomogenous solution, take y = y1 + y2, with y1 = a e^(-3x) and y2 = b e^(-x). Then

y''' = -27 y1 - y2
2y'' = 18 y1 + 2 y2
-5y' = 15 y1 + 5 y2
-6y = -6y1 + -6y2

Turns out that the left hand side is zero, so we must pick a = b = 0. (This is an exeptional situation!)

To this solution we can add homogenous solutions, i.e. of the equation y''' + 2y'' - 5y' - 6y = 0. These will be a linear combination of exponential functions, possibly with complex coefficients in the exponent. Take y = e^(px), then

y''' + 2y'' - 5y' - 6y = (p^3 + 2p^2 - 5p - 6) y = 0,
so p^3 + 2p^2 - 5p - 6 = 0. A factoring is
(p - 2)(p - 3)(p + 1) = 0, so p = 2, -1 or -3. Therefore, the functions e^(2x), e^(-x) and e^(-3x) are all solutions, as is any linear combination:

y = A e^(2x) + B e^(-x) + C e^(-3x)

for any A, B, C.