C4 Differential Equations Help?!

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?

(edited 7 years ago)

Reply 1

Kevin De Bruyne

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.

Reply 2

Shumaya

What have you done so far?

Reply 3

RetroSpectro

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

Reply 4

RetroSpectro

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

(edited 7 years ago)

Reply 5

jonnyiw

separate the variables, write 1/e^x as e^-x, (and the same for y), and then integrate both sides.

Reply 6

RetroSpectro

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

Reply 7

username2130115

You can get the particular solution by substituting the (0,1) boundary conditions into the general solution:

(edited 7 years ago)

Reply 8

Shumaya

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

Reply 9

username2130115

To express Y in terms of X, take the natural logarithm of the LHS and RHS

Reply 10

RetroSpectro

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

Reply 11

Shumaya

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

Reply 12

RetroSpectro

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

Reply 13

Shumaya

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

Reply 14

RetroSpectro

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?

Reply 15

Shumaya

Original post by RetroSpectro

@Zacken

Reply 16

Zacken

Original post by Shumaya

@Zacken

What?

Reply 17

Shumaya

Original post by Zacken

What?

Can you help him? (I can't)

Reply 18

Farhan.Hanif93

Original post by RetroSpectro

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.