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

The solution for y^{''}+4y^'-5y=2e^{3x} is y=c_{1}e^x+c_{2}e^{-5x}+1/8 e^{3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y^'-5y=2e^{3x}

y^{^{''}} + 4 y^{^{'}} - 5 y = 2 e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{8} e^{3 x}

Solution steps

y^{''} (x) + 4 y^{'} (x) - 5 y = 2 e^{3 x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{8} e^{3 x}

y = c_{1} e^{x} + c_{2} e^{- 5 x} + \frac{1}{8} e^{3 x}

4 y^{^{''}} - 4 y^{^{'}} + y = \frac{4 x}{e ^{\frac{x}{2}}}

y^{^{''}} + 2 y^{^{'}} + y = \frac{1}{6} e^{- x}

y^{^{''}} - 6 y^{^{'}} + 9 y = 12 e^{x} sin (x)

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

2 y^{^{''}} + y^{^{'}} = 4 x, y (0) = 7, y^{^{'}} (0) = - 10

What is the solution for y^{''}+4y^'-5y=2e^{3x} ?
The solution for y^{''}+4y^'-5y=2e^{3x} is y=c_{1}e^x+c_{2}e^{-5x}+1/8 e^{3x}