What is the solution for y^{''}-3y=0,y(0)=13,y^'(0)=0 ?

The solution for y^{''}-3y=0,y(0)=13,y^'(0)=0 is y=(13-13/2)e^{sqrt(3)t}+13/2 e^{-sqrt(3)t}

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y^{''}-3y=0,y(0)=13,y^'(0)=0

y^{^{''}} - 3 y = 0, y (0) = 13, y^{^{'}} (0) = 0

y = (13 - \frac{13}{2}) e^{3 t} + \frac{13}{2} e^{- 3 t}

Solution steps

y^{''} (t) - 3 y = 0

Solve linear ODE:

y = (13 - \frac{13}{2}) e^{3 t} + \frac{13}{2} e^{- 3 t}

y = (13 - \frac{13}{2}) e^{3 t} + \frac{13}{2} e^{- 3 t}

27 y^{^{''}} + (15 - 95) y^{^{'}} - 55 y = 0

324 y^{^{''}} - 612 y^{^{'}} + 289 y = 0, y (0) = 8, y^{^{'}} (0) = 5

y^{^{''}} - 13 y - 40 y = 0

25 y^{^{''}} + 10 y^{^{'}} + 5 y = 0

y^{^{''}} + 10 y^{^{'}} = 0, y (0) = - 2, y^{^{'}} (0) = - 2

What is the solution for y^{''}-3y=0,y(0)=13,y^'(0)=0 ?
The solution for y^{''}-3y=0,y(0)=13,y^'(0)=0 is y=(13-13/2)e^{sqrt(3)t}+13/2 e^{-sqrt(3)t}