# Second Order DE

• Sep 28th 2009, 04:32 AM
latavee
Second Order DE
y = c1e4x + c2e-x is the general solution of the differential equation y'' - 3y' - 4y = 0 on (-infinity, infinity). Find a member of the family that is a solution with initial conditions y(0) = 1 and y'(0) = 2.
What is the value of c1:
What is the value of c2:

So...a=1, b=3and c=-4

The discriminat equals 0, so will use -b/2a for m1 and m2

So where do I go from here to find c1 and c2? We briefly touched this in class and I was trying to figure it out myself.
• Sep 28th 2009, 05:11 AM
Prove It
Quote:

Originally Posted by latavee
y = c1e4x + c2e-x is the general solution of the differential equation y'' - 3y' - 4y = 0 on (-infinity, infinity). Find a member of the family that is a solution with initial conditions y(0) = 1 and y'(0) = 2.
What is the value of c1:
What is the value of c2:

So...a=1, b=3and c=-4

The discriminat equals 0, so will use -b/2a for m1 and m2

So where do I go from here to find c1 and c2? We briefly touched this in class and I was trying to figure it out myself.

If $\displaystyle y''(x) - 3y'(x) - 4y(x) = 0$, the characteristic equation is

$\displaystyle m^2 - 3m - 4 = 0$

$\displaystyle (m - 4)(m + 1) = 0$

$\displaystyle m = 4$ or $\displaystyle m = -1$.

Thus the solution is of the form

$\displaystyle y(x) = C_1e^{4x} + C_2e^{-x}$.

You should also be able to see that

$\displaystyle y'(x) = 4C_1e^{4x} - C_2e^{-x}$.

Using your initial conditions $\displaystyle y(0) = 1$ and $\displaystyle y'(0) = 2$, we get

$\displaystyle 1 = C_1 + C_2$ and

$\displaystyle 2 = 4C_1 - C_2$.

Solving these equations simultaneously gives

$\displaystyle 3 = 5C_1$

$\displaystyle C_1 = \frac{3}{5}$.

$\displaystyle 1 = \frac{3}{5} + C_2$

$\displaystyle C_2 = \frac{2}{5}$.

Therefore, the solution is

$\displaystyle y(x) = \frac{3}{5}e^{4x} - \frac{2}{5}e^{-x}$.
• Sep 28th 2009, 05:21 AM
latavee
Thank you
I totally had that characteristic equation not in my head! This makes sense now! Thanks!