What is the solution for x(x+1)y^'+xy=1,y(e)=1 ?

The solution for x(x+1)y^'+xy=1,y(e)=1 is y=(ln(x))/(x+1)+e/(x+1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x(x+1)y^'+xy=1,y(e)=1

x (x + 1) y^{^{'}} + x y = 1, y (e) = 1

y = \frac{ln ( x )}{x + 1} + \frac{e}{x + 1}

Solution steps

x (x + 1) y^{'} (x) + x y = 1

Solve linear ODE:

y = \frac{ln ( x )}{x + 1} + \frac{e}{x + 1}

y = \frac{ln ( x )}{x + 1} + \frac{e}{x + 1}

\int (- \frac{2 x y}{( x ^{2} + y ^{2} )}) d y

\frac{\partial}{\partial x} ((1 + x) sin (y))

\frac{d}{d x} (ln (a x a + x))

\int \frac{1}{( x ^{2} - 49 ) ^{\frac{3}{2}}} d x

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

What is the solution for x(x+1)y^'+xy=1,y(e)=1 ?
The solution for x(x+1)y^'+xy=1,y(e)=1 is y=(ln(x))/(x+1)+e/(x+1)