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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

limit as t approaches 0 of sin(8t)

t \to 0 lim (sin (8 t))

(xydy)/(dx)+y^2=2x

\frac{x y d y}{d x} + y^{2} = 2 x

derivative of sin(e^{4x})derivative of

sin (e^{4 x})

derivative of (1-xe^x/(x+e^x))

\frac{d}{d x} (\frac{1 - x e ^{x}}{x + e ^{x}})

area y=e^x,y=x,x=0,x=2 area

y = e^{x}, y = x, x = 0, x = 2

Frequently Asked Questions (FAQ)

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