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Popular >

y^'=ryln(k/y)

Solution

y^{^{'}} = ry ln (\frac{k}{y})

Solution

y = \frac{k}{e ^{e^{- t r - c_{1}}}}

Solution steps

y^{'} (t) = ry (t) ln (\frac{k}{y})

Solve separable ODE:

y = \frac{k}{e ^{e^{- t r - c_{1}}}}

y = \frac{k}{e ^{e^{- t r - c_{1}}}}

Graph

Popular Examples

limit as x approaches+5+of sqrt(x-5)

x \to + 5 + lim (x - 5)

(\partial)/(\partial x)(ln(6x^2+y^2+4))

\frac{\partial}{\partial x} (ln (6 x^{2} + y^{2} + 4))

tangent of y=x^2-4x+4,(3,1)tangent of

y = x^{2} - 4 x + 4, (3, 1)

xy^'+2y^'-(x+2)cos(x)+y=0

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

sum from n=0 to infinity of 2(-1/2)^n

n = 0 \sum \infty 2 (- \frac{1}{2})^{n}

Frequently Asked Questions (FAQ)

What is the solution for y^'=ryln(k/y) ?
The solution for y^'=ryln(k/y) is y= k/(e^{e^{-tr-c_{1)}}}