What is the solution for y^{''}+2y^'+5y=1+e^x ?

The solution for y^{''}+2y^'+5y=1+e^x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/5+1/8 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+2y^'+5y=1+e^x

y^{^{''}} + 2 y^{^{'}} + 5 y = 1 + e^{x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{5} + \frac{1}{8} e^{x}

Solution steps

y^{''} (x) + 2 y^{'} (x) + 5 y = 1 + e^{x}

Solve linear ODE:

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{5} + \frac{1}{8} e^{x}

y = e^{- x} (c_{1} cos (2 x) + c_{2} sin (2 x)) + \frac{1}{5} + \frac{1}{8} e^{x}

y^{^{''}} + 4 y = sin (t) H (t - 2 π)

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{( e ^{3 x} )}{x ^{2}}

4 y^{^{''}} + 16 y = 1

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

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

What is the solution for y^{''}+2y^'+5y=1+e^x ?
The solution for y^{''}+2y^'+5y=1+e^x is y=e^{-x}(c_{1}cos(2x)+c_{2}sin(2x))+1/5+1/8 e^x