What is the solution for y^{''}-4y^'+20y=0 ?

The solution for y^{''}-4y^'+20y=0 is y=e^{2t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-4y^'+20y=0

y^{^{''}} - 4 y^{^{'}} + 20 y = 0

y = e^{2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

f (x) = (x - 1) (x + 1)^{3}

\frac{d}{d x} (2 x tanh (x))

- \frac{9}{( 6 t + 8 ) ^{\frac{3}{2}}}

x \to 1 - lim (\frac{4}{x ^{3} - 1})

\frac{d}{d x} (3 (x)^{3} + cos (x))

What is the solution for y^{''}-4y^'+20y=0 ?
The solution for y^{''}-4y^'+20y=0 is y=e^{2t}(c_{1}cos(4t)+c_{2}sin(4t))