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Popular >

y^'=xe^{x+y}

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

derivative of x^3-3x^2-9x+4

\frac{d}{d x} (x^{3} - 3 x^{2} - 9 x + 4)

inverse oflaplace s^{-3/2}inverse laplace

s^{- \frac{3}{2}}

integral of x^2-1

\int x^{2} - 1 d x

slope of (-2.5)(-1.2)slope

(- 2.5) (- 1.2)

derivative of y=ln(xsqrt(x^2-4))derivative of

y = ln (x x^{2} - 4)

Frequently Asked Questions (FAQ)

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