What is the solution for y^{''}+9y^'=e^x,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+9y^'=e^x,y(0)=0,y^'(0)=0 is y=-1/90-1/10+1/90 e^{-9x}+1/10 e^x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}+9y^'=e^x,y(0)=0,y^'(0)=0

y^{^{''}} + 9 y^{^{'}} = e^{x}, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{1}{90} - \frac{1}{10} + \frac{1}{90} e^{- 9 x} + \frac{1}{10} e^{x}

Solution steps

y^{''} (x) + 9 y^{'} (x) = e^{x}

Solve linear ODE:

y = - \frac{1}{90} - \frac{1}{10} + \frac{1}{90} e^{- 9 x} + \frac{1}{10} e^{x}

y = - \frac{1}{90} - \frac{1}{10} + \frac{1}{90} e^{- 9 x} + \frac{1}{10} e^{x}

What is the solution for y^{''}+9y^'=e^x,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+9y^'=e^x,y(0)=0,y^'(0)=0 is y=-1/90-1/10+1/90 e^{-9x}+1/10 e^x