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Popular >

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

Solution

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

Solution

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

Solution steps

(1 - x) 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

y^{^{'}} = 2 y - 3

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

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

(1-y)e^yy^'+(y^2)/(xln(x))=0

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

2sqrt(x)(dy)/(dx)=cos^2(y),y(0)=2

2 x \frac{d y}{d x} = cos^{2} (y), y (0) = 2

(dy)/(dx)=e^{5x-5y}

\frac{d y}{d x} = e^{5 x - 5 y}

Frequently Asked Questions (FAQ)

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