What is the solution for 16y^{''}-8y^'+y=0 ?

The solution for 16y^{''}-8y^'+y=0 is y=c_{1}e^{t/4}+c_{2}e^{t/4}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

16y^{''}-8y^'+y=0

16 y^{^{''}} - 8 y^{^{'}} + y = 0

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{\frac{t}{4}} t

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{\frac{t}{4}} t

y = c_{1} e^{\frac{t}{4}} + c_{2} e^{\frac{t}{4}} t

What is the solution for 16y^{''}-8y^'+y=0 ?
The solution for 16y^{''}-8y^'+y=0 is y=c_{1}e^{t/4}+c_{2}e^{t/4}t