What is the solution for y^'+(4y)/x+4/x =0,y(1)=-2 ?

The solution for y^'+(4y)/x+4/x =0,y(1)=-2 is y=-1-1/(x^4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+(4y)/x+4/x =0,y(1)=-2

y^{^{'}} + \frac{4 y}{x} + \frac{4}{x} = 0, y (1) = - 2

y = - 1 - \frac{1}{x ^{4}}

Solution steps

y^{'} (x) + \frac{4 y}{x} + \frac{4}{x} = 0

Solve linear ODE:

y = - 1 - \frac{1}{x ^{4}}

y = - 1 - \frac{1}{x ^{4}}

\frac{d}{d x} (3 \cdot e^{\frac{x ^{3}}{9}})

x \to 0 lim (\frac{x}{x})

\int (9 x - 2)^{- 3} d x

\int coth (y) d y

\int (x^{6} + e^{3 x}) d x

What is the solution for y^'+(4y)/x+4/x =0,y(1)=-2 ?
The solution for y^'+(4y)/x+4/x =0,y(1)=-2 is y=-1-1/(x^4)