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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-2y^'-3y=e^x

y^{^{''}} - 2 y^{^{'}} - 3 y = e^{x}

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} + y = e^{x} (x^{2} - x + 1)

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

x^{^{''}} + 4 x^{^{'}} + 40 x = 4 e^{- 2 t} cos (6 t)

9 y^{^{''}} + 30 y^{^{'}} + 25 y = 2 x + 5

y^{^{''}} + 11 y^{^{'}} + 5 y = 48 e^{2 x}

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