What is the solution for x^2y^'+4xy-e^x=0 ?

The solution for x^2y^'+4xy-e^x=0 is y=(e^x)/(x^2)-(2(e^xx-e^x))/(x^4)+(c_{1})/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'+4xy-e^x=0

x^{2} y^{^{'}} + 4 x y - e^{x} = 0

y = \frac{e ^{x}}{x ^{2}} - \frac{2 ( e ^{x} x - e ^{x} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

Solution steps

x^{2} y^{'} (x) + 4 x y - e^{x} = 0

Solve linear ODE:

y = \frac{e ^{x}}{x ^{2}} - \frac{2 ( e ^{x} x - e ^{x} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

y = \frac{e ^{x}}{x ^{2}} - \frac{2 ( e ^{x} x - e ^{x} )}{x ^{4}} + \frac{c _{1}}{x ^{4}}

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

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

\int (x^{2} + 3 x - 1) e^{2 x} d x

\int \frac{sin ( 2 x ) \cdot sinh ( 2 x )}{2} d x

\int cos (x) (7 + 4 sin^{2} (x)) d x

What is the solution for x^2y^'+4xy-e^x=0 ?
The solution for x^2y^'+4xy-e^x=0 is y=(e^x)/(x^2)-(2(e^xx-e^x))/(x^4)+(c_{1})/(x^4)