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Popular >

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

Solution

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

Solution

y = - \frac{2}{- ln ( - 1 + 2 x ) + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{2}{- ln ( - 1 + 2 x ) + c _{1}}

y = - \frac{2}{- ln ( - 1 + 2 x ) + c _{1}}

Graph

Popular Examples

y^{''}+9y^'+14y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 9 y^{^{'}} + 14 y = 0, y (0) = 1, y^{^{'}} (0) = 0

derivative of x^2+5 derivative of

x^{2} + 5

(d^2y)/(dt^2)+5(dy)/(dt)=e^{-t}

\frac{d ^{2} y}{d t ^{2}} + 5 \frac{d y}{d t} = e^{- t}

5 y^{^{''}} + 35 y = 0

limit as x approaches 4+of 1

x \to 4 + lim (1)

Frequently Asked Questions (FAQ)

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