What is the solution for 2y^{''}-2y^'+13y=0 ?

The solution for 2y^{''}-2y^'+13y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-2y^'+13y=0

2 y^{^{''}} - 2 y^{^{'}} + 13 y = 0

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

Solution steps

2 y^{''} (t) - 2 y^{'} (t) + 13 y = 0

Solve linear ODE:

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

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

\frac{d}{d x} (- 4 sin^{3} (3 x))

\int \frac{1}{5000 x - x ^{2}} d x

\int_{0}^{9} 2 x + 8 - x d x

\int_{0}^{\frac{π}{4}} 1 - 4 sin^{2} (x) d x

\int 2 u sin (u) d u

What is the solution for 2y^{''}-2y^'+13y=0 ?
The solution for 2y^{''}-2y^'+13y=0 is y=e^{t/2}(c_{1}cos((5t)/2)+c_{2}sin((5t)/2))