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

The solution for x(x+1)y^'+xy=1 is y=(ln(x))/(x+1)+(c_{1})/(x+1)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

\int_{0}^{1} (arcsin (x))^{2} d x

\int \frac{1}{x ^{2} x ^{2} - 16} d x

x \to 0 lim (sin (\frac{π}{x}))

x + 2, at x = 4

\int (6 x^{2} - 5 x + 4) d x

What is the solution for x(x+1)y^'+xy=1 ?
The solution for x(x+1)y^'+xy=1 is y=(ln(x))/(x+1)+(c_{1})/(x+1)