What is the solution for 2y^{''}-2y^'+4y=0 ?

The solution for 2y^{''}-2y^'+4y=0 is y=e^{t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-2y^'+4y=0

2 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

Solution steps

2 y^{''} (t) - 2 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y = e^{\frac{t}{2}} (c_{1} cos (\frac{7 t}{2}) + c_{2} sin (\frac{7 t}{2}))

y^{^{''}} + 4 y^{^{'}} - 5 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

y^{^{''}} + y = 0, y (π) = 1, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 10 x = 0, x^{^{'}} (0) = 1, x (0) = 0

x^{^{''}} + 6 x^{^{'}} + 10 x = 0

y^{^{''}} + 4 y^{^{'}} + 7 y = 0, y (0) = 3, y^{^{'}} (0) = 7

What is the solution for 2y^{''}-2y^'+4y=0 ?
The solution for 2y^{''}-2y^'+4y=0 is y=e^{t/2}(c_{1}cos((sqrt(7)t)/2)+c_{2}sin((sqrt(7)t)/2))