What is the solution for y^{''}+4y^'+3y=4e^x,y(0)=0,y^'(0)=2 ?

The solution for y^{''}+4y^'+3y=4e^x,y(0)=0,y^'(0)=2 is y= 1/2 e^x-1/2 e^{-3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'+3y=4e^x,y(0)=0,y^'(0)=2

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

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

Solution steps

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

Solve linear ODE:

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

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

\int 3 x^{2} (x^{3} + 1)^{2} d x

\frac{\partial}{\partial x} (cos (y) + y e^{x})

\int \frac{16}{x ^{2} 16 - x ^{2}} d x

y^{^{'}} = - \frac{y}{( x + \frac{2}{y} )}

y^{^{'}} + y = b

What is the solution for y^{''}+4y^'+3y=4e^x,y(0)=0,y^'(0)=2 ?
The solution for y^{''}+4y^'+3y=4e^x,y(0)=0,y^'(0)=2 is y= 1/2 e^x-1/2 e^{-3x}