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Popular >

y^'=ae^{-by}

Solution

y^{^{'}} = a \cdot e^{- b \cdot y}

Solution

y = \frac{ln ( a t b + c _{1} b )}{b}

Solution steps

y^{'} (t) = a \cdot e^{- b \cdot y}

Solve separable ODE:

y = \frac{ln ( a t b + c _{1} b )}{b}

y = \frac{ln ( a t b + c _{1} b )}{b}

Graph

Popular Examples

(dy)/(dx)= y/((1+x))

\frac{d y}{d x} = \frac{y}{( 1 + x )}

- 2 y^{^{'}} - 3 y = 0

dy+(x^2-1)ydx=0

d y + (x^{2} - 1) y d x = 0

(2xy^2+x/(y^2))dx+4x^2ydy=0

(2 x y^{2} + \frac{x}{y ^{2}}) d x + 4 x^{2} y d y = 0

sin(y)cos(x)y^'=-cos(y)sin(x)

sin (y) cos (x) y^{^{'}} = - cos (y) sin (x)

Frequently Asked Questions (FAQ)

What is the solution for y^'=a*e^{-b*y} ?
The solution for y^'=a*e^{-b*y} is y=(ln(atb+c_{1}b))/b