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

The solution for y^'+1/x y-e^{-x^2}=0,y(1)=0 is y=-(e^{-x^2})/(2x)+1/(2ex)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+1/x y-e^{-x^2}=0,y(1)=0

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

y = - \frac{e ^{- x^{2}}}{2 x} + \frac{1}{2 e x}

Solution steps

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

Solve linear ODE:

y = - \frac{e ^{- x^{2}}}{2 x} + \frac{1}{2 e x}

y = - \frac{e ^{- x^{2}}}{2 x} + \frac{1}{2 e x}

n = 0 \sum \infty 2^{n} n^{2} x^{n}

\int ln (3) e^{x} d x

\int \frac{x ^{4} + 9}{x ^{4} + 9 x ^{2}} d x

\frac{1}{( s + 6 ) ( s - 4 )}

\frac{d}{d x} (4^{(8 x^{4} + c o s (x))^{3}})

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