What is the solution for 2y^{''}+6y^'+1=95 ?

The solution for 2y^{''}+6y^'+1=95 is y=c_{1}+c_{2}e^{-3t}+(47t)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2y^{''}+6y^'+1=95

2 y^{^{''}} + 6 y^{^{'}} + 1 = 95

y = c_{1} + c_{2} e^{- 3 t} + \frac{47 t}{3}

Solution steps

2 y^{''} (t) + 6 y^{'} (t) + 1 = 95

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} + \frac{47 t}{3}

y = c_{1} + c_{2} e^{- 3 t} + \frac{47 t}{3}

x^{^{''}} + 2 x^{^{'}} + 41 x = 11 cos (6 t)

y^{^{''}} + 3 y = 9

y^{^{''}} + y^{^{'}} = sin (x) + cos (x)

y^{^{'''}} + y^{^{''}} + 4 y^{^{'}} + 4 y = 4 x + 4

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

What is the solution for 2y^{''}+6y^'+1=95 ?
The solution for 2y^{''}+6y^'+1=95 is y=c_{1}+c_{2}e^{-3t}+(47t)/3