e^(x-y)=-x+C where y(2)=2
so I found C=3?
y=_____________
so I thought it would be y=-ln(-x+3)-x but that is wrong please help!
since y(2) = 2,
e^(2-2) = -2+c
1 = -2+c
3 = c
So far, so good.
e^(x-y) = -x+3
x-y = ln(-x+3)
y = x-ln(-x+3)
Somehow you got an extra "-" sign in there
Thank you!
To solve the given differential equation, you can use the method of separation of variables. Here's how you can do it step by step:
Step 1: Begin with the given differential equation: e^(x-y) = -x + C.
Step 2: Rearrange the equation to isolate the dependent variable y. Start by taking the natural logarithm (ln) of both sides:
ln(e^(x-y)) = ln(-x + C).
Simplifying the left side using the logarithmic property (ln(e^a) = a), we get:
x - y = ln(-x + C).
Step 3: Rearrange the equation to solve for y. Move the x term to the right side:
-y = ln(-x + C) - x.
Multiply both sides by -1 to isolate y:
y = -ln(-x + C) + x.
Step 4: Now you need to determine the value of C by using the initial condition y(2) = 2. Substitute x = 2 and y = 2 into the equation:
2 = -ln(-2 + C) + 2.
Step 5: Simplify the equation and solve for C. Start by subtracting 2 from both sides:
-2 = -ln(-2 + C).
Next, take the exponential function (e) of both sides to eliminate the logarithm:
e^(-2) = -2 + C.
Using the property (e^(ln(a)) = a), we have:
1/e^2 = -2 + C.
Adding 2 to both sides:
C = 1/e^2 + 2.
Step 6: Substitute the value of C back into the equation:
y = -ln(-x + 1/e^2 + 2) + x.
Therefore, the correct expression for y is:
y = -ln(-x + 1/e^2 + 2) + x.
Please ensure that you follow the steps and calculations carefully to arrive at the correct result.