What is the solution for y^{''}-8y^'+32y=0 ?

The solution for y^{''}-8y^'+32y=0 is y=e^{4t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-8y^'+32y=0

y^{^{''}} - 8 y^{^{'}} + 32 y = 0

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

Solution steps

y^{''} (t) - 8 y^{'} (t) + 32 y = 0

Solve linear ODE:

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

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

5 x (x - 1)

\int 2.4

x \to 0 lim (\frac{( ln ( x ) )}{x})

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

sin (4 x)

What is the solution for y^{''}-8y^'+32y=0 ?
The solution for y^{''}-8y^'+32y=0 is y=e^{4t}(c_{1}cos(4t)+c_{2}sin(4t))