What is the solution for y^{''}-4y^'+12y=0 ?

The solution for y^{''}-4y^'+12y=0 is y=e^{2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))

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y^{''}-4y^'+12y=0

y^{^{''}} - 4 y^{^{'}} + 12 y = 0

y = e^{2 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

y = e^{2 t} (c_{1} cos (2^{1 + \frac{1}{2}} t) + c_{2} sin (2^{1 + \frac{1}{2}} t))

18 y^{^{''}} + 72 y = 0, y (0) = 1, y^{^{'}} (0) = - 6

6 y^{^{''}} - 10 y^{^{'}} + 4 y = 0

5 y^{^{''}} + \frac{15}{0.2} y = 0, y (0) = 0, y^{^{'}} (0) = - 10

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 7 y = 0

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} - 15 y = 0

What is the solution for y^{''}-4y^'+12y=0 ?
The solution for y^{''}-4y^'+12y=0 is y=e^{2t}(c_{1}cos(2^{1+1/2}t)+c_{2}sin(2^{1+1/2}t))