What is the solution for 4y^{''}-4y+y=0 ?

The solution for 4y^{''}-4y+y=0 is y=c_{1}e^{(sqrt(3)t)/2}+c_{2}e^{-(sqrt(3)t)/2}

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

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

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y = c_{1} e^{\frac{3 t}{2}} + c_{2} e^{- \frac{3 t}{2}}

y^{^{'''}} + 2 y^{^{''}} - 63 y^{^{'}} = 0

2 y^{^{''}} + 2 y + y = 0, y (0) = 1, y^{^{'}} (0) = 1

0.05 q^{^{''}} + 2 q^{^{'}} + \frac{1}{0.01} q = 0

x^{^{''}} + 6 x^{^{'}} + 25 x = 0, x (0) = 3, x^{^{'}} (0) = 8

\frac{d ^{2} y}{d x ^{2}} = k y

What is the solution for 4y^{''}-4y+y=0 ?
The solution for 4y^{''}-4y+y=0 is y=c_{1}e^{(sqrt(3)t)/2}+c_{2}e^{-(sqrt(3)t)/2}