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

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

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} e^{x} + c_{2} e^{- x} + \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{4}

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 2 y^{^{'}} + 10 y = x^{2} e^{- x} cos (3 x)

y^{^{''}} + 4 y = 9 t, y (0) = 0, y^{^{'}} (0) = 7

5 y^{^{''}} - 15 y = 10 e^{x}

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

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