What is the solution for y^{''}-9y=0,y(0)=2,y^'(0)=0 ?

The solution for y^{''}-9y=0,y(0)=2,y^'(0)=0 is y=e^{3t}+e^{-3t}

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y^{''}-9y=0,y(0)=2,y^'(0)=0

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 0

y = e^{3 t} + e^{- 3 t}

Solution steps

y^{''} (t) - 9 y = 0

Solve linear ODE:

y = e^{3 t} + e^{- 3 t}

y = e^{3 t} + e^{- 3 t}

25 y^{^{''}} - y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 13 y = 0

8 y^{^{''}} + 12 y^{^{'}} + 5 y = 0

6 \frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} - y = 0

y^{^{''}} + 9 y = 0, y (0) = sin (3 x)

What is the solution for y^{''}-9y=0,y(0)=2,y^'(0)=0 ?
The solution for y^{''}-9y=0,y(0)=2,y^'(0)=0 is y=e^{3t}+e^{-3t}