What is the solution for y^{''}-6y^'+45y=0 ?

The solution for y^{''}-6y^'+45y=0 is y=e^{3t}(c_{1}cos(6t)+c_{2}sin(6t))

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y^{''}-6y^'+45y=0

y^{^{''}} - 6 y^{^{'}} + 45 y = 0

y = e^{3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 45 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

\int_{0}^{1} \frac{53}{x ^{5}} d x

\frac{\partial}{\partial x} (- \frac{x}{y})

\int_{- 1}^{1} (2 x - 1)^{2} d x

\int sin^{2} (14 x) d x

\frac{d y}{d x} = - 5 - 5 x + y

What is the solution for y^{''}-6y^'+45y=0 ?
The solution for y^{''}-6y^'+45y=0 is y=e^{3t}(c_{1}cos(6t)+c_{2}sin(6t))