Second Order DE

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September 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
using the quad equation.... 4+/-Sqrt16-16/2a

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.
September 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
using the quad equation.... 4+/-Sqrt16-16/2a

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

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

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

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

Thus the solution is of the form

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

You should also be able to see that

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

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

$1 = C_1 + C_2$ and

$2 = 4C_1 - C_2$ .

Solving these equations simultaneously gives

$3 = 5C_1$

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

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

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

Therefore, the solution is

$y(x) = \frac{3}{5}e^{4x} - \frac{2}{5}e^{-x}$ .
September 28th 2009, 05:21 AM
latavee

Thank you

I totally had that characteristic equation not in my head! This makes sense now! Thanks!