What is the solution for y^{''}-4a^2y^'+y=0 ?

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

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y^{''}-4a^2y^'+y=0

y^{^{''}} - 4 a^{2} y^{^{'}} + y = 0

y = c_{1} e^{(2 a^{2} + 4 a^{4} - 1) t} + c_{2} e^{(2 a^{2} - 4 a^{4} - 1) t}

Solution steps

y^{''} (t) - 4 a^{2} y^{'} (t) + y = 0

Solve linear ODE:

y = c_{1} e^{(2 a^{2} + 4 a^{4} - 1) t} + c_{2} e^{(2 a^{2} - 4 a^{4} - 1) t}

y = c_{1} e^{(2 a^{2} + 4 a^{4} - 1) t} + c_{2} e^{(2 a^{2} - 4 a^{4} - 1) t}

y^{^{''}} + 36 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y^{^{''}} + 8 y^{^{'}} = 0, y (0) = 3, y^{^{'}} (0) = 16

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = \frac{3}{2}

\frac{1}{2} y^{^{''}} - \frac{1}{3} y^{^{'}} - 2 y = 0

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

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