What is the solution for y^{''}-12y^'+38y=0 ?

The solution for y^{''}-12y^'+38y=0 is y=e^{6t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-12y^'+38y=0

y^{^{''}} - 12 y^{^{'}} + 38 y = 0

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

Solution steps

y^{''} (t) - 12 y^{'} (t) + 38 y = 0

Solve linear ODE:

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

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

\int \frac{1}{2} cos^{2} (3 θ) d θ

\int r^{2} (r^{3} - 4)^{7} d r

\frac{d y}{d x} + 2 y = 2 x

y^{^{'}} + \frac{1}{x} y = x y^{2}

x y + x

What is the solution for y^{''}-12y^'+38y=0 ?
The solution for y^{''}-12y^'+38y=0 is y=e^{6t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))