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Popular >

(y^2+y)-xy^'=0

Solution

(y^{2} + y) - x y^{^{'}} = 0

Solution

y = - \frac{x}{x - c _{1}}

Solution steps

(y^{2} + y) - x y^{'} (x) = 0

Solve separable ODE:

y = - \frac{x}{x - c _{1}}

y = - \frac{x}{x - c _{1}}

Popular Examples

x^3+2y(dy)/(dx)=0

x^{3} + 2 y \frac{d y}{d x} = 0

(df)/(dx)+f(x)^2=0

\frac{df}{d x} + f (x)^{2} = 0

(dy)/(dx)=(yln(y))/(cot(x))

\frac{d y}{d x} = \frac{y ln ( y )}{cot ( x )}

y(4+x^2)dy-x(2+y^2)dx=0

y (4 + x^{2}) d y - x (2 + y^{2}) d x = 0

(dy)/(dx)xln(x)-y=0,y(2)=ln(4)

\frac{d y}{d x} x ln (x) - y = 0, y (2) = ln (4)

Frequently Asked Questions (FAQ)

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