What is the solution for 4y^{''}-21y^'+20y=0,y(0)=0 ?

The solution for 4y^{''}-21y^'+20y=0,y(0)=0 is y=-c_{2}e^{4t}+c_{2}e^{(5t)/4}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

4y^{''}-21y^'+20y=0,y(0)=0

4 y^{^{''}} - 21 y^{^{'}} + 20 y = 0, y (0) = 0

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

Solution steps

4 y^{''} (t) - 21 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

y = - c_{2} e^{4 t} + c_{2} e^{\frac{5 t}{4}}

What is the solution for 4y^{''}-21y^'+20y=0,y(0)=0 ?
The solution for 4y^{''}-21y^'+20y=0,y(0)=0 is y=-c_{2}e^{4t}+c_{2}e^{(5t)/4}