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Popular >

y^'=k(50-y)

Solution

y^{^{'}} = k (50 - y)

Solution

y = - e^{- k t - c_{1}} + 50

Solution steps

y^{'} (t) = k (50 - y)

Solve separable ODE:

y = - e^{- k t - c_{1}} + 50

y = - e^{- k t - c_{1}} + 50

Graph

Popular Examples

(dy)/(dx)=ky-10000

\frac{d y}{d x} = k y - 10000

(dy)/(dx)=-ln|x|y

\frac{d y}{d x} = - ln ∣ x ∣ y

ln(y^6)(dy)/(dx)=x^6y,y(1)=e^2

ln (y^{6}) \frac{d y}{d x} = x^{6} y, y (1) = e^{2}

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

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

y^'=(1+y^2)^{0.5}

y^{^{'}} = (1 + y^{2})^{0.5}

Frequently Asked Questions (FAQ)

What is the solution for y^'=k(50-y) ?
The solution for y^'=k(50-y) is y=-e^{-kt-c_{1}}+50