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