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

The solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 is y=-1/16 e^{-x}-(e^{-x}x)/4+1/16 e^{3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

Solution steps

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

Solve linear ODE:

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

y = - \frac{1}{16} e^{- x} - \frac{e ^{- x} x}{4} + \frac{1}{16} e^{3 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 9 x^{2}

y^{^{''}} + 6 y^{^{'}} - 7 y = 21 x^{2} - 8 x - 30

y^{^{''}} + y^{^{'}} - 2 y = e^{- 2 \cdot x} sin (x)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x + 4, y (0) = 1, y^{^{'}} (0) = - 3

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

What is the solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 ?
The solution for y^{''}+2y^'+y=e^{3x},y(0)=0,y^'(0)=0 is y=-1/16 e^{-x}-(e^{-x}x)/4+1/16 e^{3x}