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Popular >

(x+1)y^'+y^2=0

Solution

(x + 1) y^{^{'}} + y^{2} = 0

Solution

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

Solution steps

(x + 1) y^{'} (x) + y^{2} = 0

Solve separable ODE:

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

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

Popular Examples

(dx)/(dt)+x/(100)=75

\frac{d x}{d t} + \frac{x}{100} = 75

(x^2-3)(dy)/(dx)=x(y^2+16),y(2)=4

(x^{2} - 3) \frac{d y}{d x} = x (y^{2} + 16), y (2) = 4

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

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

y^'=(4x)/(1+2y),y(9)=0

y^{^{'}} = \frac{4 x}{1 + 2 y}, y (9) = 0

\frac{d x}{x} = \frac{d y}{y}

Frequently Asked Questions (FAQ)

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