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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

y = c_{1} e^{5 x} + c_{2} e^{- x} - \frac{1}{4} e^{3 x}

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