What is the solution for y^{''}-6y^'+18y=0 ?

The solution for y^{''}-6y^'+18y=0 is y=e^{3t}(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^{''}-6y^'+18y=0

y^{^{''}} - 6 y^{^{'}} + 18 y = 0

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

Solution steps

y^{''} (t) - 6 y^{'} (t) + 18 y = 0

Solve linear ODE:

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

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

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

\frac{2}{s + 1}

\int (3 x + 4)^{\frac{5}{2}} d x

\int_{0}^{2} (2 x - x^{2}) d x

x \to 1 + lim (\frac{x - 1}{x + 1})

What is the solution for y^{''}-6y^'+18y=0 ?
The solution for y^{''}-6y^'+18y=0 is y=e^{3t}(c_{1}cos(3t)+c_{2}sin(3t))