Final answer to the problem
$\ln\left|e^y-1\right|-y=x+C_0$
Got another answer? Verify it here!
Step-by-step Solution
How should I solve this problem?
- Choose an option
- Exact Differential Equation
- Linear Differential Equation
- Separable Differential Equation
- Homogeneous Differential Equation
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Integrate by substitution
- Integrate by parts
- Load more...
Can't find a method? Tell us so we can add it.
1
Rewrite the differential equation using Leibniz notation
$\frac{dy}{dx}=e^y-1$
Learn how to solve problems step by step online.
$\frac{dy}{dx}=e^y-1$
Learn how to solve problems step by step online. Solve the differential equation y^'=e^y-1. Rewrite the differential equation using Leibniz notation. Group the terms of the differential equation. Move the terms of the y variable to the left side, and the terms of the x variable to the right side of the equality. Integrate both sides of the differential equation, the left side with respect to y, and the right side with respect to x. Solve the integral \int\frac{1}{e^y-1}dy and replace the result in the differential equation.
Final answer to the problem
$\ln\left|e^y-1\right|-y=x+C_0$
