# differential equation??

### 2 Answers

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- nyphdinmdLv 74 weeks ago
I assume you want the solution?

3y" + 2y' + y = 0

let y = e^(rx) then the ODE transforms into

(3r^2 + 2r + 1)e^(rx) =0 = (3r^2 + 2r + 1)

Solve the quadratic for r:

r = -1/3 +/- (1/6)*sqrt((2)^2 -4*3) = (1/3)*{ -1 +/- (1/2) sqrt(-2)} =(1/3){ -1 +/- i/sqrt(2)}

So

y = e^(-x/3)*{A*e^(i x/(3*sqrt(2))) + B*e^(-i x/(3*sqrt(2))) }

A and B are determined from boundary or initial conditions

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