# Initial value problem

1. Sep 28, 2009

### -EquinoX-

1. The problem statement, all variables and given/known data
How can I find a general solution to this problem:

5y" + 9y' - 9y = 0

2. Relevant equations

3. The attempt at a solution

I am confused with second order different equations in it

2. Sep 28, 2009

### CompuChip

Usually with this type of equations (homogeneous ordinary differential equations) the easiest way is to plug in a trial solution $y(t) = e^{\lambda t}$, then solve the resulting polynomial for $\lambda$ and make a linear combination to get the general solution.

(For non-homogeneous equations, add to that a particular solution).

3. Sep 28, 2009

### -EquinoX-

can't I just integrate twice and get the answer? how can I solve this using the method of undeterminant coefficients

4. Sep 28, 2009

### Staff: Mentor

No. You can integrate y'' twice, but how are you going to integrate y' and y twice?

CompuChip has given you some good advice. Try it out.

5. Sep 28, 2009

### -EquinoX-

I don't get what he meant by make a linear combination

6. Sep 28, 2009

### Staff: Mentor

Assume that the solution is of the form y = ert. Calculate the first and second derivatives of this function and plug them into your differential equation. After you factor out ert, you will have a quadratic that you can solve for r. (CompuChip used lambda, but the idea is the same.) You should get two values for r--r1 and r2, meaning that y1 = er1t and y2 = er2t are solutions of this DE. The general solution is all possible linear combinations of these two functions. IOW, y = c1er1t + c2er2t.

A linear combination of functions is the sum of multiples of those functions.

Clear?

7. Sep 30, 2009

### -EquinoX-

what if it only has one root, say (y-3)^2, so the only root is y = 3

Last edited: Sep 30, 2009
8. Sep 30, 2009

### Staff: Mentor

Then your solutions would be e3t and te3t. Your general solution would be the linear combinations of these two functions.

IOW, yg = c1e3t + c2te3t.

9. Sep 30, 2009

### -EquinoX-

if it's y^2 + 25y , then the two roots are y = 0 and y = -25. so yg = c1 + c2e^(-25t).

10. Oct 1, 2009

### CompuChip

Yes. You can understand why you only get one solution which is non-trivial (i.e. not equal to a constant number everywhere): if r = 0 is a solution to the polynomial, then that means that your equation must have been of the form
a y'' + b y' = 0
so without a term in y.
In that case you can view it as a (simpler) first order equation for v = y':
a v' + b v = 0
which of course only has one independent solution.

In general, if your equation is nth order (i.e. the highest derivative of y that occurs is the nth derivative) you get n independent solutions which you should add (indeed, when you plug in the exponential solution, you get a polynomial of degree n). You should be advised, however, that the roots of the polynomial are not always real, like for example in y'' + y = 0.

11. Oct 1, 2009

### -EquinoX-

well if have initial value y(0) = 2 and y'(0)= 5

then what I get is

y' = -25c2e^(-25t)
5 = -25c2
so c2 = -1/5

and c1 is equal to 11/5

is the true? reason I asked is because the stupid webassign won't accept this answer

12. Oct 1, 2009