Find the particular solution of differential equations that satisfies the initial condition

dx/dy=e^(x+y) x(1)=0

I know you have to get all the y and dy on one side and x and dx on the other so you can intagrate and stuff, but im having trouble getting there

dx/dy = e^(x+y)

dx/dy = e^x * e^y
e^-x dx = e^y dy
-e^-x = e^y + c

x(1) = 0, so
0 = e+c
c = -e

-e^-x = e^y - e
e^-x = e - e^y
x = -ln(e - e^y)

Thanks so much!!

To find the particular solution of the given differential equation, we will use the method of separation of variables.

The given differential equation is:
dx/dy = e^(x+y)

To begin, we separate the variables by moving the terms involving x to one side and the terms involving y to the other side:
1/e^(x+y) dx = dy

Next, we integrate both sides with respect to their respective variables. Integrating the left side:
∫ 1/e^(x+y) dx = ∫ dy

Integrating the left side requires a little manipulation. We can rewrite 1/e^(x+y) as e^(-x-y) since the integral of e^u du is just e^u. Therefore, the left side becomes:
∫ e^(-x-y) dx = ∫ dy

Now, we can integrate both sides:
-∫ (e^(-x-y)) dx = y + C

To integrate the left side, consider that we have e^(-x-y) with respect to x only. Treating y as a constant, integration gives us:
- e^(-x-y) = y + C

Now, we solve for y in terms of x and the constant C:
y = -e^(-x-y) - C

Using the initial condition x(1) = 0, we substitute x = 1 and y = 0 into the equation:
0 = -e^(-1-0) - C

Solving for C:
C = -1

Therefore, the particular solution of the given differential equation that satisfies the initial condition x(1) = 0 is:
y = -e^(-x-y) - 1

To find the particular solution of the given differential equation, you need to separate variables and integrate. Here's how you can do it step by step:

1. Start with the given differential equation:
dx/dy = e^(x+y)

2. Rearrange the equation so that all instances of x and dx are on one side, and all instances of y and dy are on the other side:
dx = e^(x+y) dy

3. Divide both sides by e^(x+y) to isolate dy on one side:
dx / e^(x+y) = dy

4. Integrate both sides with respect to their respective variables:
∫ dx / e^(x+y) = ∫ dy

5. Evaluate the integrals:
∫ dx / e^(x+y) = ∫ dy
=> ln|e^(x+y)| = y + C (where C is the constant of integration)

6. Simplify the natural logarithm using the property ln(e^a) = a:
ln(e^(x+y)) = y + C
=> x+y = y + C

7. Solve for x:
x = C

8. To find the value of the constant C, substitute the initial condition x(1) = 0:
x(1) = 0
=> C = 0

9. Finally, substitute C back into the equation x = C:
x = 0

Therefore, the particular solution of the differential equation that satisfies the initial condition x(1) = 0 is x = 0.