What is the solution for 49y^{''}+28y^'+4y=0 ?

The solution for 49y^{''}+28y^'+4y=0 is y=c_{1}e^{-(2t)/7}+c_{2}e^{-(2t)/7}t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

49y^{''}+28y^'+4y=0

49 y^{^{''}} + 28 y^{^{'}} + 4 y = 0

y = c_{1} e^{- \frac{2 t}{7}} + c_{2} e^{- \frac{2 t}{7}} t

Solution steps

49 y^{''} (t) + 28 y^{'} (t) + 4 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{2 t}{7}} + c_{2} e^{- \frac{2 t}{7}} t

y = c_{1} e^{- \frac{2 t}{7}} + c_{2} e^{- \frac{2 t}{7}} t

16 y^{^{''}} - 40 y^{^{'}} + 25 y = 0, y (0) = 3, y^{^{'}} (0) = - 94

x \to 0 lim (x \cdot sin (\frac{π}{x}))

m = \frac{9}{5 + h} - \frac{9}{5}

n = 1 \sum \infty e^{- n} + 1

\frac{d}{d x} (1 + \frac{1}{4 x})

What is the solution for 49y^{''}+28y^'+4y=0 ?
The solution for 49y^{''}+28y^'+4y=0 is y=c_{1}e^{-(2t)/7}+c_{2}e^{-(2t)/7}t