# A question on second order linear equations

1. Jun 23, 2014

### mech-eng

Hi, all. While solving a second order linear differential equation why do we have to use linear independent but two solutions. For example, when solving y''- y = 0 , y(0) = 5 and y'(0) = 3 , we use ex and e-x and then we write y = c1*ex+ c2*e-x-

2. Jun 23, 2014

### maajdl

That's because any linear combination of a solution of the DE is a solution of the DE.
Initial conditions restrict the set of possible solutions.

3. Jun 23, 2014

### maajdl

That's because any linear combination of a solution of the DE is a solution of the DE.
Initial conditions restrict the set of possible solutions.

4. Jun 23, 2014

### mech-eng

I am especially interested in the number "two", "two solutions". Why we derive general solution
from any two?

5. Jun 23, 2014

### SteamKing

Staff Emeritus
You seem to be confused about the solutions to ODEs.

For the equation y" - y = 0, with y(0) = 5 and y'(0) = 3, we have a homogeneous ODE for which we find the complementary solution by assuming y = e^(rx). Differentiating and substituting into the original ODE gives:

r^2 * e^(rx) - e^rx = 0

Dividing both sides by e^rx (which is always > 0, for any x), we obtain the characteristic polynomial

r^2 - 1 = 0 or r^2 = 1, which has solutions r = 1 and r = -1.

This means that e^x and e^-x are both solutions to the original ODE. Since the original ODE is linear, the sum of the individual solutions obtained from solving the characteristic polynomial must also be a solution, thus

y = C1*e^-x + C2 * e^x

By applying the initial conditions y(0) = 5 and y'(0) = 3, the coefficients of the general solution C1 and C2 can be determined.

6. Jun 23, 2014

### mech-eng

How do we guess that complementary solution? In my example you did it as erx

7. Jun 23, 2014

### SteamKing

Staff Emeritus
That's the standard technique for solving linear ODEs with constant coefficients. See:

http://en.wikipedia.org/wiki/Linear_differential_equation

Section: Homogeneous equations with constant coefficients