What is the solution for y^'=y+x^2e^x+1-1/e ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=y+x^2e^x+1-1/e

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (3 x y^{2} + 2 y cos (x))

\int \frac{1}{x + x x} d x

\int 8 x + \frac{1}{x} d x

\frac{d}{d x} (\frac{x + 4}{3 x ^{2} e ^{x}})

\int_{0}^{1} e^{x} d x

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