Question

find a solution to the initial value problem
Y" +Y' = 0 y(pi/3) = 2, y'(pi/3) = -4

Answer 1

\(y(x)=A\sin(x)+B\cos(x)\)
\(y'(x)=A\cos(x)-B\sin(x)\)
\(y(\pi/3)=A\sin(\pi/3)+B\cos(\pi/3)=2\)
\(y'(\pi/3)=A\cos(\pi/3)iB\sin(\pi/3)=-4\)
Solve the last 2 equations for A and B

Answer 2

Sorry, the last equation should be:
y′(π/3)=Acos(π/3)-Bsin(π/3)=−4

Answer 3

I have to use the method concerning characteristic equation and the general solution,
\[y = c1e^(t \alpha) \cos(\beta t) + c2e^(t \alpha)\sin(\beta t)... where... r = \alpha + or - \beta i\]

Answer 4

nevermind, I made a copy error and the second y is not primed...already figured it out. Thanks for help tho!