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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

limit as x approaches 5+of ln(sin(x))

x \to 5 + lim (ln (sin (x)))

y^'=((xy+2y))/(x+3)

y^{^{'}} = \frac{( x y + 2 y )}{x + 3}

expand (2x+x^2)/((1+x)^2)expand

\frac{2 x + x ^{2}}{( 1 + x ) ^{2}}

(\partial)/(\partial x)(ln(1+3x^2y^3))

\frac{\partial}{\partial x} (ln (1 + 3 x^{2} y^{3}))

limit as x approaches 3 of (x^2-9)/(x+3)

x \to 3 lim (\frac{x ^{2} - 9}{x + 3})

Frequently Asked Questions (FAQ)

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