e and ln problem

[Jason76]

Is all of this right?

\(\displaystyle -dy + y = \ln(x - y) + x - dx\)

\(\displaystyle e^{-dy + y} = e^{\ln(x - y) + x - dx}\)

\(\displaystyle e^{-dy} * e^{ y} = x - y * e^{x} * e^{-dx}\)

\(\displaystyle e^{-dy} * e^{ y} - y = x - y - y * e^{x} * e^{-dx}\)

\(\displaystyle e^{-dy} * e^{ y} - y = x * e^{x} * e^{-dx}\)

\(\displaystyle \ln(e^{-dy}) * \ln( e^{ y}) - \ln(y) = \ln(x) * \ln(e^{x}) * \ln(e^{-dx})\)

\(\displaystyle -dy + y - \ln(y) = \ln(x) + x -dx\)

 

[Jason76]

ln and Diff. Equations

How to solve this last step?

\(\displaystyle y + \ln(y) + dy = -\ln(x) - x + dx\)

\(\displaystyle \int y + \ln(y) + dy = \int -\ln(x) - x + dx\)

\(\displaystyle u = \ln x\)

\(\displaystyle u' = \dfrac{1}{x}\)

\(\displaystyle v = x\)

\(\displaystyle v' = dx\)

Formula: \(\displaystyle \int uv' = uv - \int vu'\)

\(\displaystyle \int \ln x dx = \ln x x -\int x \dfrac{1}{x} dx\)

\(\displaystyle = \ln x x - \dfrac{x^{2}}{2}\ln x + C\) What's on the left hand side?

One way is to do integration by parts to the ln expressions on both sides, but what about the other variables?

 

[daon2]

I'm sorry but I have no idea what you are doing. You cannot integrate "y + lny + dy", it is not a valid differential.

 

[Jason76]

In response to Daon, it came from this problem

\(\displaystyle \dfrac{dx - dy}{x - y} = \ln(x - y)\)

\(\displaystyle (x - y)\dfrac{dx - dy}{x - y} = \ln(x - y) (x - y)\)

\(\displaystyle dx - dy = \ln(x - y) + x - y\)

\(\displaystyle dx - dx - dy = \ln(x - y) + x - y - dx\)

\(\displaystyle -dy = \ln(x - y) + x - y - dx\)

\(\displaystyle -dy + y = \ln(x - y) + x - y + y - dx\)

\(\displaystyle -dy + y = \ln(x - y) + x - dx\)

\(\displaystyle e^{-dy + y} = e^{\ln(x - y) + x - dx}\)

\(\displaystyle e^{-dy} * e^{ y} = x - y * e^{x} * e^{-dx}\)

\(\displaystyle e^{-dy} * e^{ y} - y = x - y - y * e^{x} * e^{-dx}\)

\(\displaystyle e^{-dy} * e^{ y} - y = x * e^{x} * e^{-dx}\)

\(\displaystyle \ln(e^{-dy}) * \ln( e^{ y}) - \ln(y) = \ln(x) * \ln(e^{x}) * \ln(e^{-dx})\)

\(\displaystyle -dy + y - \ln(y) = \ln(x) + x -dx\)

\(\displaystyle (-1)(-dy + y -\ln(y)) = (-1)(\ln(x) + x -dx)\)


Below this is what we were dealing with:

\(\displaystyle dy - y + \ln(y) = -\ln(x) - x + dx\)

\(\displaystyle y + \ln(y) + dy = -\ln(x) - x + dx\)

\(\displaystyle \int y + \ln(y) + dy = \int -\ln(x) - x + dx\)

 

[daon2]

You seem not to understand what I meant when I said this problem doesn't make sense, and cannot be solved. Assuming y is a function of x, it is equivalent to solving:

\(\displaystyle \dfrac{1-y'}{x-y} dx = \ln(x-y)\)

On the left we have a differential. On the right we have a function.

I believe the original poster of this question was asked to show the antiderivative of the left-hand side is equal to the right-hand side and did not copy the problem down correctly.

 

[Jason76]

I think your right. I tried to solve the problem by trying to get rid of the ln by using e, and then separating variables. However, as you said, it was messed up from the start.

 

[Deleted member 4993]

Is all of this right?

\(\displaystyle -dy + y = \ln(x - y) + x - dx\)

\(\displaystyle e^{-dy + y} = e^{\ln(x - y) + x - dx}\)

\(\displaystyle e^{-dy} * e^{ y} = x - y * e^{x} * e^{-dx}\) ...................... This line does not follow from above

\(\displaystyle e^{-dy} * e^{ y} - y = x - y - y * e^{x} * e^{-dx}\)

\(\displaystyle e^{-dy} * e^{ y} - y = x * e^{x} * e^{-dx}\) ...................... This line does not follow from above

\(\displaystyle \ln(e^{-dy}) * \ln( e^{ y}) - \ln(y) = \ln(x) * \ln(e^{x}) * \ln(e^{-dx})\)

\(\displaystyle -dy + y - \ln(y) = \ln(x) + x -dx\)

What are you exactly trying to do?

 

[lookagain]

Assuming y is a function of x, it is equivalent to solving:

> > \(\displaystyle \dfrac{1-y'}{x-y} dx = \ln(x-y)\) < <

On the left we have a differential. On the right we have a function.

I believe the original poster of this question was asked to show the antiderivative of the left-hand side is equal to the right-hand side and did not copy the problem down correctly.

If, instead, someone were to take the derivative of the quantity on the right side of this incorrect equation, I see it equaling the quantity on the left side, but without the "dx."