What is the solution for 2y^{''}-3y^'+2y=0 ?

The solution for 2y^{''}-3y^'+2y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}-3y^'+2y=0

2 y^{^{''}} - 3 y^{^{'}} + 2 y = 0

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

Solution steps

2 y^{''} (t) - 3 y^{'} (t) + 2 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

y = e^{\frac{3 t}{4}} (c_{1} cos (\frac{7 t}{4}) + c_{2} sin (\frac{7 t}{4}))

64 y^{^{'''}} + 112 y^{^{''}} + 49 y^{^{'}} = 0

y^{^{''}} = 5 y, y^{0} = - 2, y^{2} = 10

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

y^{^{''}} - 2 y^{^{'}} + 5 y = 0, y (0) = - 1, y^{^{'}} (0) = 3

\frac{d ^{4} y}{d t ^{4}} - y = 0

What is the solution for 2y^{''}-3y^'+2y=0 ?
The solution for 2y^{''}-3y^'+2y=0 is y=e^{(3t)/4}(c_{1}cos((sqrt(7)t)/4)+c_{2}sin((sqrt(7)t)/4))