how to solve: dy/dx=e^(x-y), y(0)=ln3

Well, before we worry about the initial condition, we should solve the DE. dy/dx = e^(x-y) If you are not certain how to solve here, consider the format we had and also the Algebra property: a^(x - y) = a^(x) a^(-y)

Thanks, I've forgotten most of my algebra.

You're welcome! :) If it may help, you can usually find quite a few resources on any given Algebra properties on google.

dy/dx=(e^x)(e^(-y)) dy/dx=F(x)G(y) dy/dx(1/G(y))=F(x) int(e^y)dx=int(e^x)dx e^y=e^x y=x correct so far?

Your process is correct, although there is one thing we need to include from the integral since it is an indefinite integral: \(\int e^{y} \; \text{d}y = \int e^{x} \; \text{d}x \) \(e^{y} = e^{x} \color{red}{+C} \) The constant of integration here.

From there, we can again take the natural log of both sides: y = ln (e^x + C) <-- we cant really fix right side here now, but it will have to do. Basically, the constant of variation is what gives a differential equation many possible solutions, and the initial value will allow us to figure out a value of C for a particular solution.

got it!

Good! :) Then, we just want to use y(0) = ln 3 to solve for C in that equation. ln 3 = ln (e^0 + C) ln 3 = ln (1 + C) 3 = 1 + C C = 2 --> y = ln (e^x + 2) is the particular solution.

In all differential equations, we will always have some constant of variation, or a few of them depending on the order of the derivative.

* have some constant of variation in the solution

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