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Popular >

y^'=e^xe^{-y}

Solution

y^{^{'}} = e^{x} e^{- y}

Solution

y = ln (e^{x} + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (e^{x} + c_{1})

y = ln (e^{x} + c_{1})

Popular Examples

(dP)/(dt)=4sqrt(Pt),P(1)=5

\frac{d P}{d t} = 4 Pt, P (1) = 5

(x^2-1)(dy)/(dx)+2xy=x

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

(1-x^2)(dy)/(dx)+xy=x

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

(dy)/(dx)=e^xsin(y)

\frac{d y}{d x} = e^{x} sin (y)

(dp)/(dt)=t^2p-3p+t^2-3

\frac{d p}{d t} = t^{2} p - 3 p + t^{2} - 3

Frequently Asked Questions (FAQ)

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