What is the solution for y^{''}=4y,y(0)=0,y^'(0)=1 ?

The solution for y^{''}=4y,y(0)=0,y^'(0)=1 is y= 1/4 e^{2t}-1/4 e^{-2t}

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y^{''}=4y,y(0)=0,y^'(0)=1

y^{^{''}} = 4 y, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{4} e^{2 t} - \frac{1}{4} e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = \frac{1}{4} e^{2 t} - \frac{1}{4} e^{- 2 t}

y = \frac{1}{4} e^{2 t} - \frac{1}{4} e^{- 2 t}

y^{^{'}} = x (x - 1)

(y + x^{3} + x y^{2}) d x - x d y = 0

\int e^{1 - x} x d x

\int (x - 1) x d x

\frac{\partial}{\partial x} (x^{2} sin (4 t))

What is the solution for y^{''}=4y,y(0)=0,y^'(0)=1 ?
The solution for y^{''}=4y,y(0)=0,y^'(0)=1 is y= 1/4 e^{2t}-1/4 e^{-2t}