What is the solution for y^{''}-12y^'+52y=0 ?

The solution for y^{''}-12y^'+52y=0 is y=e^{6t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-12y^'+52y=0

y^{^{''}} - 12 y^{^{'}} + 52 y = 0

y = e^{6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 12 y^{'} (t) + 52 y = 0

Solve linear ODE:

y = e^{6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{6 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\int_{0}^{1} (1 - x)^{9} d x

f (x) = 7 x + 9

f (x) = 18 x^{2} - 27 x

\frac{\partial}{\partial x} ((x^{3} - y^{2})^{2})

\frac{\partial}{\partial x} ((x + y)^{5})

What is the solution for y^{''}-12y^'+52y=0 ?
The solution for y^{''}-12y^'+52y=0 is y=e^{6t}(c_{1}cos(4t)+c_{2}sin(4t))