What is the solution for y^{''}-6y^'+10y=0 ?

The solution for y^{''}-6y^'+10y=0 is y=e^{3t}(c_{1}cos(t)+c_{2}sin(t))

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y^{''}-6y^'+10y=0

y^{^{''}} - 6 y^{^{'}} + 10 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 10 y = 0

Solve linear ODE:

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

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

C (x) = \frac{4 x + 50}{x + 2}

\int x e^{11 x + 3} d x

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

\int 4 arcsin (x) d x

x \to 1 - lim (x^{2} - 3)

What is the solution for y^{''}-6y^'+10y=0 ?
The solution for y^{''}-6y^'+10y=0 is y=e^{3t}(c_{1}cos(t)+c_{2}sin(t))