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Popular >

y^'=y^2(1+e^x)

Solution

y^{^{'}} = y^{2} (1 + e^{x})

Solution

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

Solution steps

y^{'} (x) = y^{2} (1 + e^{x})

Solve separable ODE:

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

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

Popular Examples

derivative of (4x^5+7/(x^3))

\frac{d}{d x} (\frac{4 x ^{5} + 7}{x ^{3}})

y^{''}-2y^'+y=((-6e^t))/(t^2+1)

y^{^{''}} - 2 y^{^{'}} + y = \frac{( - 6 e ^{t} )}{t ^{2} + 1}

tangent of f(x)=x^3-6x^2-34x+40,\at x=2 tangent of

f (x) = x^{3} - 6 x^{2} - 34 x + 40, at x = 2

derivative of f(x)=x^2arcsin(3x)derivative of

f (x) = x^{2} arcsin (3 x)

f (x) = x^{5} e^{x}

Frequently Asked Questions (FAQ)

What is the solution for y^'=y^2(1+e^x) ?
The solution for y^'=y^2(1+e^x) is y=-1/(x+e^x+c_{1)}