What is the solution for xy^'+2y=(3e^{-4x})/x ?

The solution for xy^'+2y=(3e^{-4x})/x is y=-(3e^{-4x})/(4x^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=(3e^{-4x})/x

x y^{^{'}} + 2 y = \frac{3 e ^{- 4 x}}{x}

y = - \frac{3 e ^{- 4 x}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

Solution steps

x y^{'} (x) + 2 y = \frac{3 e ^{- 4 x}}{x}

Solve linear ODE:

y = - \frac{3 e ^{- 4 x}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

y = - \frac{3 e ^{- 4 x}}{4 x ^{2}} + \frac{c _{1}}{x ^{2}}

x \to - 1 + lim (\frac{x}{x ^{2} - 1})

\int_{0}^{4} 5 e^{- x^{2}} d x

\frac{cot ( x )}{x}

x \to 0 lim (\frac{20 sin ( x )}{4})

x \to 0 lim (x (e^{x} - 1) ln (x))

What is the solution for xy^'+2y=(3e^{-4x})/x ?
The solution for xy^'+2y=(3e^{-4x})/x is y=-(3e^{-4x})/(4x^2)+(c_{1})/(x^2)