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Popular >

y^'=x(y-1)

Solution

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

Solution

y = e^{\frac{x ^{2}}{2} + c_{1}} + 1

Solution steps

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

Solve separable ODE:

y = e^{\frac{x ^{2}}{2} + c_{1}} + 1

y = e^{\frac{x ^{2}}{2} + c_{1}} + 1

Popular Examples

(dy)/(dt)+4y=5,y(0)=1

\frac{d y}{d t} + 4 y = 5, y (0) = 1

(dy)/(dx)=c_{1}y+1

\frac{d y}{d x} = c_{1} y + 1

(dv)/(dt)=g-kv(t),v=0,t=0

\frac{d v}{d t} = g - k v (t), v = 0, t = 0

(dy)/(dx)+2y=11

\frac{d y}{d x} + 2 y = 11

(dy)/(dx)y=e^{x^{2ln(x)}}

\frac{d y}{d x} y = e^{x^{2 l n (x)}}

Frequently Asked Questions (FAQ)

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