What is the solution for y^{''}-4y^'+6y=0 ?

The solution for y^{''}-4y^'+6y=0 is y=e^{2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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

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

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

Solution steps

y^{''} (t) - 4 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{'''}} - 9 y^{^{''}} + 28 y^{^{'}} - 20 y = 0

2 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

y^{^{''''}} + k^{2} y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} - k^{2} y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

What is the solution for y^{''}-4y^'+6y=0 ?
The solution for y^{''}-4y^'+6y=0 is y=e^{2t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))