What is the solution for y^{''}-12y^'+61y=0 ?

The solution for y^{''}-12y^'+61y=0 is y=e^{6t}(c_{1}cos(5t)+c_{2}sin(5t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-12y^'+61y=0

y^{^{''}} - 12 y^{^{'}} + 61 y = 0

y = e^{6 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) - 12 y^{'} (t) + 61 y = 0

Solve linear ODE:

y = e^{6 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{6 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

\frac{d y}{d x} = \frac{y}{x} ln (y)

\int_{0}^{\frac{π}{2}} - 6 sin^{2} (3 x) d x

\frac{\partial}{\partial x} (2 x y (9 - x - y))

x^{2} y^{2}

\int 2 x y e^{x^{2} y} d x

What is the solution for y^{''}-12y^'+61y=0 ?
The solution for y^{''}-12y^'+61y=0 is y=e^{6t}(c_{1}cos(5t)+c_{2}sin(5t))