What is the solution for y^{''}+8y^'+25y=0 ?

The solution for y^{''}+8y^'+25y=0 is y=e^{-4t}(c_{1}cos(3t)+c_{2}sin(3t))

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y^{''}+8y^'+25y=0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0

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

Solution steps

y^{''} (t) + 8 y^{'} (t) + 25 y = 0

Solve linear ODE:

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

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

y = x^{3} - 2 x + 1, (2, 5)

\int_{- 2}^{1} ∣ x ∣ d x

\frac{d y}{d x} + a y = b

\int \frac{x ^{3}}{( 7 - x ^{4} ) ^{5}} d x

What is the solution for y^{''}+8y^'+25y=0 ?
The solution for y^{''}+8y^'+25y=0 is y=e^{-4t}(c_{1}cos(3t)+c_{2}sin(3t))