What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=5 ?

The solution for y^{''}-4y=0,y(0)=0,y^'(0)=5 is y= 5/4 e^{2t}-5/4 e^{-2t}

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y^{''}-4y=0,y(0)=0,y^'(0)=5

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 5

y = \frac{5}{4} e^{2 t} - \frac{5}{4} e^{- 2 t}

Solution steps

y^{''} (t) - 4 y = 0

Solve linear ODE:

y = \frac{5}{4} e^{2 t} - \frac{5}{4} e^{- 2 t}

y = \frac{5}{4} e^{2 t} - \frac{5}{4} e^{- 2 t}

8 y^{^{''}} + c y^{^{'}} + 128 y = 0, y (0) = 5, y^{^{'}} (0) = 8

y^{^{''}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 6

y^{^{''}} - \frac{4 y ^{^{'}}}{3} + \frac{97 y}{36} = 0

y^{^{''}} - 10 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = \frac{31}{2}

4 \frac{d y}{d x} - 3 \frac{d ^{2} y}{d x ^{2}} = 0

What is the solution for y^{''}-4y=0,y(0)=0,y^'(0)=5 ?
The solution for y^{''}-4y=0,y(0)=0,y^'(0)=5 is y= 5/4 e^{2t}-5/4 e^{-2t}