What is the solution for y^{''}+4y^'+y=2 ?

The solution for y^{''}+4y^'+y=2 is y=c_{1}e^{(-2+sqrt(3))t}+c_{2}e^{(-2-sqrt(3))t}+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+y=2

y^{^{''}} + 4 y^{^{'}} + y = 2

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} + 2

Solution steps

y^{''} (t) + 4 y^{'} (t) + y = 2

Solve linear ODE:

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} + 2

y = c_{1} e^{(- 2 + 3) t} + c_{2} e^{(- 2 - 3) t} + 2

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 t e^{5 t}

y^{^{''}} - 6 y^{^{'}} + 8 y = - 3 e^{2 x} - 5 x e^{- 4 x} + 3 x

y^{^{''''}} - y^{^{''}} = 3 x^{2} - 2 x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1

\frac{1}{6} y^{^{''}} + 6 y = tan (6 x) - \frac{1}{3} e^{3 x}

What is the solution for y^{''}+4y^'+y=2 ?
The solution for y^{''}+4y^'+y=2 is y=c_{1}e^{(-2+sqrt(3))t}+c_{2}e^{(-2-sqrt(3))t}+2