What is the solution for xy^'=xe^x-2y ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

xy^'=xe^x-2y

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

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

Solution steps

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

Solve linear ODE:

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

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

x \to - 6 lim (x^{2})

4 + 2 x + 10 x e^{x}

f (x) = (3 - 5 x e^{x}) (3 x + 2)

\int_{1}^{5} (x^{2} - x + 1) d x

\frac{d}{d x} (- sin (4 x))

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