What is the solution for xy^'+2y=2e^x ?

The solution for xy^'+2y=2e^x is y=(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^'+2y=2e^x

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

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

Solution steps

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

Solve linear ODE:

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

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

\frac{d x}{d t} = sin (x)

\int e^{\frac{x}{16}} d x

\int x^{2} (x - 7) d x

\int x (2 x^{2} + 1)^{6} d x

t + 1

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