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Popular >

y^'=e^y,y(0)=0

Solution

y^{^{'}} = e^{y}, y (0) = 0

Solution

y = - ln (- t + 1)

Solution steps

y^{'} (t) = e^{y}

Solve separable ODE:

y = - ln (- t + 1)

y = - ln (- t + 1)

Graph

Popular Examples

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2 x^{3} H (x) (x)^{2} d x + x^{4} H (x) d H (x) = 0

(1+x^2)*y^'=x(1-y)

(1 + x^{2}) \cdot y^{^{'}} = x (1 - y)

(dy)/(dx)-2xy=4x,y(0)=3

\frac{d y}{d x} - 2 x y = 4 x, y (0) = 3

(dy)/(dx)=y*sin(x)

\frac{d y}{d x} = y \cdot sin (x)

y(x^3dy+y^3dx)=0

y (x^{3} d y + y^{3} d x) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^y,y(0)=0 ?
The solution for y^'=e^y,y(0)=0 is y=-ln(-t+1)