What is the solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)=1 ?

The solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)=1 is y=2e^{t/2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)=1

4 y^{^{''}} - 8 y^{^{'}} + 3 y = 0, y (0) = 2, y^{^{'}} (0) = 1

y = 2 e^{\frac{t}{2}}

Solution steps

4 y^{''} (t) - 8 y^{'} (t) + 3 y = 0

Solve linear ODE:

y = 2 e^{\frac{t}{2}}

y = 2 e^{\frac{t}{2}}

2 y^{^{''}} + 4 y^{^{'}} + 8 y = 0, y (0) = 5, y^{^{'}} (0) = 0

2 y^{^{''}} + y = 0, y (0) = e^{- \frac{x}{2}}

4 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

(D^{3} - 2 D^{2} - D + 2) y = 0

2 y^{^{'''}} + y^{^{''}} - 8 y^{^{'}} + 5 y = 0

What is the solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)=1 ?
The solution for 4y^{''}-8y^'+3y=0,y(0)=2,y^'(0)=1 is y=2e^{t/2}