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Popular Calculus >

(1+e^x)yy^'=e^x,y(0)=1

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

y^'=0.05y(1-y/(4000))

(dy)/(dx)y=sin^3(cos(9x))

(dy)/(dx)=85yx^{16},y=4

1/x (dy)/(dx)=2-4y

Frequently Asked Questions (FAQ)

What is the solution for (1+e^x)*y*y^'=e^x,y(0)=1 ?
The solution for (1+e^x)*y*y^'=e^x,y(0)=1 is y=sqrt(2ln(1+e^x)+1-2ln(2))