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

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

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 = c_{1} cos (\frac{t}{2}) + c_{2} sin (\frac{t}{2})

Solution steps

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

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 10 \frac{d y}{d x} - 3 y = 0

y^{^{''}} + 15 y = 0, y (0) = sin (k t)

y^{^{''}} - \frac{1}{c} y = 0

2 y^{^{''}} + 12 y = 0

y^{^{''}} - 2 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 4

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