What is the solution for y^{''}-6y^'+25y=0 ?

The solution for y^{''}-6y^'+25y=0 is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}-6y^'+25y=0

y^{^{''}} - 6 y^{^{'}} + 25 y = 0

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\frac{\partial}{\partial y} (x sin (6 x^{2} y))

\int 24 sec^{5} (x) d x

\frac{d y}{d x} = \frac{( y ^{2} - 1 )}{( x ^{2} - 1 )}, y (0) = 1

\frac{\partial}{\partial x} (4 x cos (x y))

f (x) = x^{2} + x + 2, 0, 12

What is the solution for y^{''}-6y^'+25y=0 ?
The solution for y^{''}-6y^'+25y=0 is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))