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Calc 2 Initital-value Problem Question?
Hello!
The question states:
Solve the initial-value problem.
y'' - 2y' + y = 0 , y(2) = 0 , y'(2) = 1
So....
r^2 - 2r + 1 = 0
Using the quadratic formula to find the values:
(- - 2 +/- sqRt((-2)^2 - 4 * 1 * 1))/(2 * 1)
2 +/- sqRt(0)/2
2/2
1
In the book, it states that this means the general solution is:
y = c1e^rx + c2e^rx, where in this case r = 1.
So then I differentiate, but....
y' = c1e^x + c2e^x = y = c1e^x + c2e^x
Because they're the same, I can't solve for the constants.
What am I doing wrong?
Thanks!!
I followed your suggestion, and used:
C₁exp(x) + C₂xexp(x)....
So I set it up as:
c1e^2 + c2 * 2 * e^2 = 0
Then I differentiated:
(e^x + x * e^x)' = c1e^x + c2e^x(x + 2)
So..
c1e^2 + c2 * 2 * e^2 = 0
c1e^2 + c2e^2(2 + 2) = c1e^2 + 4 * c2e^2 = 1... multiplied this whole equation by -1
c1e^2 - c1e^2 + c2 * 2 * e^2 - 4 * c2e^2 = -1
-2 * c2 * e^2 = -1
c2 = 1/2e^2
c1e^2 + 1/2e^2 * 2 * e^2 = 0
c1e^2 = -1/2e^2 * 2 * e^2
c1e^2 = -1
c1 = -1/e^2
So the answer is:
-e^x/e^2 + xe^x/2e^2
But this isn't right.
Why?
1 Answer
- δοτζοLv 78 years agoFavourite answer
The solutions must be linearly independent for the Superposition Principle to apply. So you need to make your solutions linearly independent by multiplying one of them by x.
y = C₁exp(x) + C₂xexp(x)
is your general solution. You can use that to solve for the constants. I'm sure your book explains this sometime before complex solutions.
