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# C4 Differential Equations Help?!

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1. Im having trouble trying to solve this differential equation from the A2 edexcel C4 text book:

Solve and Express Y in terms of x

dx/dy = e^x / e^y

and is given that x=0 when y=1

What i have so far: (-e^-x) = (-e^-y) + (e^-1) - 1

any thoughts?
2. (Original post by RetroSpectro)
Im having trouble trying to solve this differential equation from the A2 edexcel C4 text book:

dx/dy = e^x / e^y

and is given that x=0 when y=1

any thoughts?
Separation of variables.
3. What have you done so far?
4. (Original post by SeanFM)
Separation of variables.
yeah I did that, I got quite far into it but just cant seem to get the answer
5. (Original post by Shumaya)
What have you done so far?
Ive tried 2 methods.

One where i used a^x = a^x/lna

and the other when i integrated it normally by recognition

I managed to get C for both methods but then struggle to figure out the next step
6. separate the variables, write 1/e^x as e^-x, (and the same for y), and then integrate both sides.
7. (Original post by jonnyiw)
separate the variables, write 1/e^x as e^-x, (and the same for y), and then integrate both sides.
Yeah I did that, sorry forgot to mention that you have to express Y in terms of x
8. You can get the particular solution by substituting the (0,1) boundary conditions into the general solution:
9. (Original post by RetroSpectro)
=I managed to get C for both methods but then struggle to figure out the next step
Write out your equation so far then we can help with the next step
10. To express Y in terms of X, take the natural logarithm of the LHS and RHS
11. (Original post by Shumaya)
Write out your equation so far then we can help with the next step
(e^-x) = (e^-y) + (e^-1) - 1
12. (Original post by RetroSpectro)
(e^-x) = (e^-y) + (e^-1) - 1
What did you get as the integral of 1/e^x? it should be -e^-x
13. (Original post by Shumaya)
What did you get as the integral of 1/e^x? it should be -e^-x
(-e^-x) = (-e^-y) + (e^-1) - 1

My bad, I forgot to put that in
14. (Original post by RetroSpectro)
(-e^-x) = (-e^-y) + (e^-1) - 1

My bad, I forgot to put that in
Now isolate the e^-y, take the natural log of both sides, and multiply both sides by -1
15. (Original post by Shumaya)
Now isolate the e^-y, take the natural log of both sides, and multiply both sides by -1
According to answers at the back of the book you should get:

y = x + 1 - ln (e - (e^x+1) + e^x)

which does not seem to be the solution according to your proposed method.

Do you think the book is incorrect?
16. (Original post by RetroSpectro)
According to answers at the back of the book you should get:

y = x + 1 - ln (e - (e^x+1) + e^x)

which does not seem to be the solution according to your proposed method.

Do you think the book is incorrect?
Zacken
17. (Original post by Shumaya)
Zacken
What?
18. (Original post by Zacken)
What?
Can you help him? (I can't)
19. (Original post by RetroSpectro)
According to answers at the back of the book you should get:

y = x + 1 - ln (e - (e^x+1) + e^x)

which does not seem to be the solution according to your proposed method.

Do you think the book is incorrect?
(Original post by Shumaya)
Can you help him? (I can't)
The book is correct. For some (rather irritating) reason, LaTeX is not compiling as it should; but following on from Shumaya's working, note that:

e^(-x) + e^(-1) - 1 = [e + e^x - e^(x+1)] * e^[-(x+1)]

Taking logs of this in conjunction with the law for the log of a product yields the desired result.
20. (Original post by Farhan.Hanif93)
The book is correct. For some (rather irritating) reason, LaTeX is not compiling as it should; but following on from Shumaya's working, note that:

e^(-x) + e^(-1) - 1 = [e + e^x - e^(x+1)] * e^[-(x+1)]

Taking logs of this in conjunction with the law for the log of a product yields the desired result.
Oh I see, makes sense now.

However, a very unusual question

Thanks

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