Kranz
- 1
- 0
Hello everyone and thanks for looking at my thread,
I had some trouble solving this ODE which was in a textbook by Henry J. Ricardo:
x(e^y - y') = 2.
This problem is from a section dealing with linear equations, but there is a hint beside the problem which reads, "Hint: Think of y as the independent variable, x as the dependent variable, and rewrite the equation in terms of dx/dy."
I tried to solve it by doing the following:
e^y - y' = 2 / x
y' = e^y - 2 / x
y' = (xe^y - 2) / x
dx/dy = x / (xe^y - 2)
xe^y + e^y (dx/dy) = x
dx/dy + x = xe^{-y}
dx/dy = x(e^{-y} - 1)
(1 / x)dx = (e^{-y} - 1)dy
ln|x| + C = -e^{-y} - y
However, I know that this is not the correct solution; I'm guessing that I went wrong somewhere when I rewrote the equation in terms of dx/dy, but I don't know how else I would approach the problem. Where did I go wrong with my work, and how would I proceed to solve the problem correctly?
Thank you very much for your help!
I had some trouble solving this ODE which was in a textbook by Henry J. Ricardo:
x(e^y - y') = 2.
This problem is from a section dealing with linear equations, but there is a hint beside the problem which reads, "Hint: Think of y as the independent variable, x as the dependent variable, and rewrite the equation in terms of dx/dy."
I tried to solve it by doing the following:
e^y - y' = 2 / x
y' = e^y - 2 / x
y' = (xe^y - 2) / x
dx/dy = x / (xe^y - 2)
xe^y + e^y (dx/dy) = x
dx/dy + x = xe^{-y}
dx/dy = x(e^{-y} - 1)
(1 / x)dx = (e^{-y} - 1)dy
ln|x| + C = -e^{-y} - y
However, I know that this is not the correct solution; I'm guessing that I went wrong somewhere when I rewrote the equation in terms of dx/dy, but I don't know how else I would approach the problem. Where did I go wrong with my work, and how would I proceed to solve the problem correctly?
Thank you very much for your help!
Last edited: