Solving ODE via Separation Method: Need Assistance

[Mugged]

Can anyone help me solve this ODE:
..in other words, find a general solution?

dy/dx = e^(x+y)

I use a separation method, but i can't take the natural log of -e^(-y).
So, help?

 

[Filip Larsen]

Perhaps you can move the minus in front of e^-y to the other side of your equation before you apply log.

 

[HallsofIvy]

Looks to me like you have forgotten the constant of integration. You should get something like $y= -ln(C- e^x)$ which exists only for $C> e^x$ or $x< ln(C)$

 

[shelovesmath]

Can anyone help me solve this ODE:
..in other words, find a general solution?

dy/dx = e^(x+y)

I use a separation method, but i can't take the natural log of -e^(-y).
So, help?

e^(x+y) = e^x * e*y
e^(-y) dy = e^x dx
integrate both sides

 

[blue_raver22]

Sure, I would be happy to assist you with solving this ODE using the separation method. First, let's review the steps for solving an ODE using separation:

1. Rewrite the ODE in the form dy/dx = f(x,y)
2. Separate the variables by moving all terms containing y to one side and all terms containing x to the other side.
3. Integrate both sides of the equation.
4. Solve for y to find the general solution.

Applying these steps to the given ODE, we have:

1. dy/dx = e^(x+y)
2. dy/e^y = e^x dx
3. Integrate both sides:
∫dy/e^y = ∫e^x dx
Using the substitution u = -y, du = -dy, this becomes:
-∫du/e^u = ∫e^x dx
-1/e^u = e^x + C
4. Solve for y:
-1 = e^(x-y) + Ce^y
Rearranging, we get:
e^(x-y) = Ce^y - 1
Taking the natural log of both sides:
x-y = ln(Ce^y - 1)
Simplifying:
y = x - ln(Ce^y - 1)
This is the general solution to the given ODE.

I hope this helps and feel free to ask for further clarification if needed.