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))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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{d}{d x} (1 + 7 sin (x) - tan (x))

y (1 + x^{2}) y^{^{'}} - x (2 + y^{2}) = 0

f (x) = - 8 x^{2} + 10, (0.5, 8)

\frac{d}{d x} ((x^{2} - 9)^{\frac{1}{2}})

n = 1 \sum \infty (- \frac{2}{3})^{n}

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))