Question

Solve: $e^{x+y}dx + dy = 0$

Answer 1

y = -\ln(e^x + K)

Answer 2

The given differential equation is: $e^{x+y}dx + dy = 0$

Step 1: Separate the variables. First, rewrite the term $e^{x+y}$ as $e^x e^y$. So the equation becomes: $e^x e^y dx + dy = 0$

Move the $e^x e^y dx$ term to the right side: $dy = -e^x e^y dx$

Now, separate the variables by dividing both sides by $e^y$ (note that $e^y$ is never zero): $\frac{dy}{e^y} = -e^x dx$ This can be written as: $e^{-y} dy = -e^x dx$

Step 2: Integrate both sides. Integrate both sides of the separated equation: $\int e^{-y} dy = \int -e^x dx$

For the left side integral: $\int e^{-y} dy = -e^{-y}$

For the right side integral: $\int -e^x dx = -e^x$

Combining the results and adding the constant of integration, $C$: $-e^{-y} = -e^x + C$

Step 3: Solve for y. Multiply the entire equation by -1 to make the terms positive, or rearrange: $e^{-y} = e^x - C$ Since $C$ is an arbitrary constant, $-C$ is also an arbitrary constant. Let's denote it as $K$ for clarity, so $K = -C$. $e^{-y} = e^x + K$

To solve for $y$, take the natural logarithm (ln) of both sides: $\ln(e^{-y}) = \ln(e^x + K)$ $-y = \ln(e^x + K)$

Finally, multiply by -1 to get $y$: $y = -\ln(e^x + K)$

Here, $K$ is an arbitrary constant of integration. For the logarithm to be defined, $e^x + K > 0$.

The final solution is: $y = -\ln(e^x + K)$