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Popular >

(1+e^x)y^'+e^xy=0

Solution

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

Solution

y = \frac{c _{1}}{1 + e ^{x}}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1}}{1 + e ^{x}}

y = \frac{c _{1}}{1 + e ^{x}}

Popular Examples

derivative of y= 1/(x^{1/2)}derivative of

y = \frac{1}{x ^{\frac{1}{2}}}

(dy)/(dx)=(1+2y^2)/(ysin(x))

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

integral of (3x-2)/(1-6x-9x^2)

\int \frac{3 x - 2}{1 - 6 x - 9 x ^{2}} d x

integral of (6x-5sqrt(x)+1)/(2sqrt(x))

\int \frac{6 x - 5 x + 1}{2 x} d x

derivative of (4x+5^3)

\frac{d}{d x} ((4 x + 5)^{3})

Frequently Asked Questions (FAQ)

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