What is the solution for y^{''}-36y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-36y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{6t}+1/2 e^{-6t}

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y^{''}-36y=0,y(0)=1,y^'(0)=0

y^{^{''}} - 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{2}) e^{6 t} + \frac{1}{2} e^{- 6 t}

Solution steps

y^{''} (t) - 36 y = 0

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{6 t} + \frac{1}{2} e^{- 6 t}

y = (1 - \frac{1}{2}) e^{6 t} + \frac{1}{2} e^{- 6 t}

4 y^{^{''}} + 51 y^{^{'}} + 36 y = 0, y (0) = - 5, y^{^{'}} (0) = - 20

y^{^{''}} + 9 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

4 y^{^{''}} - 4 y^{^{'}} + 17 y = 0

4 y^{^{''}} - 2 y^{^{'}} + 9 y = 0

y^{^{''}} + 19 y^{^{'}} + 90 y = 0

What is the solution for y^{''}-36y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-36y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{6t}+1/2 e^{-6t}