What is the solution for y^{''}-25y=0,y(0)=1,y^'(0)=0 ?

The solution for y^{''}-25y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{5t}+1/2 e^{-5t}

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y^{''}-25y=0,y(0)=1,y^'(0)=0

y^{^{''}} - 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- 5 t}

Solution steps

y^{''} (t) - 25 y = 0

Solve linear ODE:

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- 5 t}

y = (1 - \frac{1}{2}) e^{5 t} + \frac{1}{2} e^{- 5 t}

What is the solution for y^{''}-25y=0,y(0)=1,y^'(0)=0 ?
The solution for y^{''}-25y=0,y(0)=1,y^'(0)=0 is y=(1-1/2)e^{5t}+1/2 e^{-5t}