What is the solution for y^{''}-y=xe^{x/2} ?

The solution for y^{''}-y=xe^{x/2} is y=c_{1}e^x+c_{2}e^{-x}+(-12e^{x/2}x-16e^{x/2})/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y=xe^{x/2}

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

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- 12 e ^{\frac{x}{2}} x - 16 e ^{\frac{x}{2}}}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- 12 e ^{\frac{x}{2}} x - 16 e ^{\frac{x}{2}}}{9}

y = c_{1} e^{x} + c_{2} e^{- x} + \frac{- 12 e ^{\frac{x}{2}} x - 16 e ^{\frac{x}{2}}}{9}

y^{^{''}} + 9 y = e^{2 x} + x^{2} sin (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x^{- 1} e^{x}

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

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t + e^{t}

x^{^{''}} + 25 x = 2 cos (5.1 t)

What is the solution for y^{''}-y=xe^{x/2} ?
The solution for y^{''}-y=xe^{x/2} is y=c_{1}e^x+c_{2}e^{-x}+(-12e^{x/2}x-16e^{x/2})/9