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

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

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

linear

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

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

Solution steps

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

Solve linear ODE:

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

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

(4 D^{5} - 15 D^{3} - 5 D^{2} + 15 D + 9) y = 0

f^{^{''}} (x) = 6 f^{^{'}} (x) - 6 f (x)

16 y^{^{''}} + 64 y = 0, y (0) = 0.3, y^{^{'}} (0) = - 0.2

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

y^{^{''}} - 5 y + 6 y = 0

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