What is the solution for y^{''}+y^'=xe^x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y^'=xe^x

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

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

Solution steps

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

Solve linear ODE:

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

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

2 y^{^{''}} - 2 y^{^{'}} - 4 y = 2 e^{3} t

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

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

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

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

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