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y''' - 5y'' + 100y' - 500y = 0; y(0) = 0, y'(0) = 10, y''(0) = 250

given that y_1(x) = e^5x is one particular solution of the differential equation.

r^3 - 5r^2 + 100r - 500 = 0

r = +/- 10i or 5

complementary solution y_c = e^5x(c1cos10x + c2sin10x)

general solution y_g = y_c + y_p = e^5x(c1cos10x + c2sin10x + 1)

y(0) = c1 + 0 + 1 = 0

c1 = -1

y'(0) = 5e^5x((-1)cos10x + c2sin10x) + e^5x(-10(-1)sin10x + 10c2cos10x) = 10

5*(-1) + 10(0) + 10c2 = 10

10c2 = 15

c2 = 3/2

y(x) = e^5x(-cos10x + (3/2)sin10x + 1)

The book's answer is 2e^5x - 2cos10x. How did they get that? And do I need y''(0) = 250? I have one real root and 2 complex roots, so

e^ax(c1cosbx + c2sinbx) should work. And I don't need y'' because I don't have a c3 to solve for, yes? What am I doing wrong?

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# Homework Help: Homogeneous Differential Equation

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