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Popular >

y^'=1+x-y-xy

Solution

y^{^{'}} = 1 + x - y - x y

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

limit as x approaches 2 of (sin(2x))/x

x \to 2 lim (\frac{sin ( 2 x )}{x})

tangent of y=sqrt(x)(81.9)tangent of

y = x (81.9)

derivative of f(x)=6cos(pi/6)derivative of

f (x) = 6 cos (\frac{π}{6})

3(1+t^2)(dy)/(dt)=2ty(y^3-1)

3 (1 + t^{2}) \frac{d y}{d t} = 2 t y (y^{3} - 1)

integral from 0 to 7 of x/(x^2-1)

\int_{0}^{7} \frac{x}{x ^{2} - 1} d x

Frequently Asked Questions (FAQ)

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