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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

tangent of f(x)=-x+5x^{-1/2}-3,(1,1)tangent of

f (x) = - x + 5 x^{- \frac{1}{2}} - 3, (1, 1)

(dy)/(dx)=((x-3))/(y-3)

\frac{d y}{d x} = \frac{( x - 3 )}{y - 3}

integral from 0 to infinity of 1/(1+x)

\int_{0}^{\infty} \frac{1}{1 + x} d x

integral of 2cos^2(θ)

\int 2 cos^{2} (θ) d θ

derivative of (e^x+e^{-x}/(e^x-e^{-x)})

\frac{d}{d x} (\frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}})

Frequently Asked Questions (FAQ)

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