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

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

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

y^{^{''}} - 4 y = 0, 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 = 0

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}

f (x) = \frac{4}{1 + x ^{2}}, (1, 2)

\frac{3 s + 4}{s ^{2} + 2}

x \to 0 lim (x^{2} + 3 x - 6)

\frac{5}{2} y^{4} y^{^{'}} - \frac{y ^{5} + 4}{x ^{3}} = 0, y (1) = 0

(- 6, - 5), (4, 4)

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