What is the solution for y^{''}+4y^'+3y=3e^{2x} ?

The solution for y^{''}+4y^'+3y=3e^{2x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/5 e^{2x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+3y=3e^{2x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 3 e^{2 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 3 e^{2 x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{1}{5} e^{2 x}

y^{^{''}} - 4 y^{^{'}} + 4 y = 4 x + 6

y^{^{''}} - 4 y = 28

y^{^{'''}} + 196 y^{^{'}} = tan (14 x)

y^{^{''}} + 4 y^{^{'}} + 16 y = 4 sin (6 t)

y^{^{''}} + 16 y = 16, y (0) = 1, y^{^{'}} (0) = 4

What is the solution for y^{''}+4y^'+3y=3e^{2x} ?
The solution for y^{''}+4y^'+3y=3e^{2x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+1/5 e^{2x}