What is the solution for y^{''}-6y^'+58y=0 ?

The solution for y^{''}-6y^'+58y=0 is y=e^{3t}(c_{1}cos(7t)+c_{2}sin(7t))

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y^{''}-6y^'+58y=0

y^{^{''}} - 6 y^{^{'}} + 58 y = 0

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

Solution steps

y^{''} (t) - 6 y^{'} (t) + 58 y = 0

Solve linear ODE:

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

y = e^{3 t} (c_{1} cos (7 t) + c_{2} sin (7 t))

\int x^{3} 4 + x^{2} d x

\int 3 x x^{2} + 8 d x

\frac{d ^{2} y}{d x ^{2}} + 10 \frac{d y}{d x} + 25 y = 0

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

\frac{d}{d x} (sin (x^{2} y^{3}))

What is the solution for y^{''}-6y^'+58y=0 ?
The solution for y^{''}-6y^'+58y=0 is y=e^{3t}(c_{1}cos(7t)+c_{2}sin(7t))