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Popular >

e^{x+y}-y^'=0

Solution

e^{x + y} - y^{^{'}} = 0

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

(dy)/(dx)=(x-2ycot(2x))

\frac{d y}{d x} = (x - 2 y cot (2 x))

integral of (7sqrt(x))/(x^3-4)

\int \frac{7 x}{x ^{3} - 4} d x

integral of ((x-4)/(x^9))

\int (\frac{x - 4}{x ^{9}}) d x

area y=(x-2)^2,y=4-x,y=2sqrt(x-2)area

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

derivative of (75/(sqrt(12+x^2+y^2)))

\frac{d}{d x} (\frac{75}{12 + x ^{2} + y ^{2}})

Frequently Asked Questions (FAQ)

What is the solution for e^{x+y}-y^'=0 ?
The solution for e^{x+y}-y^'=0 is y=-ln(-e^x-c_{1})