What is the solution for y^{''}-12y^'+45y=0 ?

The solution for y^{''}-12y^'+45y=0 is y=e^{6t}(c_{1}cos(3t)+c_{2}sin(3t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-12y^'+45y=0

y^{^{''}} - 12 y^{^{'}} + 45 y = 0

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

Solution steps

y^{''} (t) - 12 y^{'} (t) + 45 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} ((x - sin (x)) (x + cos (x)))

2 x \frac{d y}{d x} + y^{3} e^{- 2 x} = 2 x y

\int e^{x} \frac{1}{x} d x

cos (x y)

\frac{\partial}{\partial x} (- 10 x e^{- 5 x^{2} - 2 y^{2}})

What is the solution for y^{''}-12y^'+45y=0 ?
The solution for y^{''}-12y^'+45y=0 is y=e^{6t}(c_{1}cos(3t)+c_{2}sin(3t))