What is the solution for y^{''}-4y^'+40y=0 ?

The solution for y^{''}-4y^'+40y=0 is y=e^{2t}(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}-4y^'+40y=0

y^{^{''}} - 4 y^{^{'}} + 40 y = 0

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) - 4 y^{'} (t) + 40 y = 0

Solve linear ODE:

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{2 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

\int (3 x + 1)^{2} - 2 d x

\frac{d}{d x} (sin (2 x + 5))

\int e^{e^{x}} e^{x} d x

\frac{d}{d x} (2. 5^{x})

y = x^{2} - x, (4, 14)

What is the solution for y^{''}-4y^'+40y=0 ?
The solution for y^{''}-4y^'+40y=0 is y=e^{2t}(c_{1}cos(6t)+c_{2}sin(6t))