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Solve the initial value problem 2y" + 5y' - 3y = 0 y(0) =7, y'(0) = 7

Can someone please help. The answer I came up with is 35 exp(-x) - 28 exp(-3x/2), but my answer is wrong.

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Hmmm. I get

y = -e^(x/2) - 7e^(-3x)

Better check your algebra.
2t^2+5t-3 = (2t-1)(t+3)

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