Need help with differential equation questions

dayflower2016

I've been revising for a maths retake using the chemistry maths book second edition, and there are some questions I can't do (for reference for those with the book; questions 21 and 22 from section 11.3 of the book).

21. when x0=2 and y0=, solve differential equation dy/dx = (y(y+1))/(x(x+1))
answer for this question should be y=x-1

22. when x0=0 and y0=0, solve dy/dx = e^(x+y)
answer should be -ln(2-e^(x))

I can't see how to get these answers out, so anyone willing to give me a thorough explanation will have my thanks!

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Reply 1

B_9710

Both are separable ODE's.
First one can be rearranged to

\displaystyle \frac{1}{y(y+1)} \ dy = \frac{1}{x(x+1)} \ dx

.

Remember for the second one,

e^{x+y} = e^x \cdot e^y

(edited 7 years ago)

Reply 2

Diraclohr

For the first, rearrange to get all of the y's on one side, all of the x's on the other. You will end up with dy*1/(y(y+1))=dx*1/(x(x+1)). Now you can integrate both sides (don't worry about the limits too much, they come out in the constant of integration). This integral is best tackled with partial fractions. You get a bunch of ln's flying about, and you can just exponentiate them all away! plugging in y0 and x0 give this constant of integration.

For the second, note that e^(x+y)=(e^x)*(e^y) then do a similar thing to before; rearrange to get the y's on one side and the x's on the other. It remains to integrate simple exponentials and deal with the constant of integration by plugging y0 and x0 in.

Hope this helps

Reply 3

dayflower2016

Thanks guys, think I've got question 21 now, didn't think of partial fractions! I'll have a go with q22, my issue is I know in theory how to do the differential equation, I just get bogged down in the integration!

Reply 4

NT226

Original post by dayflower2016

For question 21, multiply both sides by dx and divide by y(y+1). You should get 1/(y(y+1))dy = 1/(x(x+1))dx
You should then be able to solve this using separation of variables. To integrate both sides, you would need to use partial fractions. Then rearrange to get y= and sub in the initial conditions.
For question 22, e^(x+y) = e^x e^y. Get all the y expressions on one side and x on the other as before. Solve again by separation of variables.

I hope this helps. If you need anything else, just ask.

Reply 5

username2640523

Original post by dayflower2016

21.use partial fractions for the 1/(x(x+1))

22. Looks like you use e^(y+x)=e^Y*e^x

Hope that helps

Edit:As the comment above says aswell^^

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(edited 7 years ago)

Reply 6

B_9710

I'm glad that people agree with me.

Reply 7

dayflower2016

Original post by NT226

I've been able to get y/(y+1) = x/(x+1) +C
where do I go from here?

Reply 8

B_9710

Original post by dayflower2016

I've been able to get y/(y+1) = x/(x+1) +C
where do I go from here?

You will have to use partial fractions to integrate

\displaystyle \frac{1}{x(x+1)}

.
Note that

\displaystyle \frac{1}{x(x+1)} \equiv \frac{1}{x} - \frac{1}{x+1}

Reply 9

dayflower2016

Original post by B_9710

You will have to use partial fractions to integrate

\displaystyle \frac{1}{x(x+1)}

.
Note that

\displaystyle \frac{1}{x(x+1)} \equiv \frac{1}{x} - \frac{1}{x+1}

The y/(y+1) = x/(x+1) +C was after integration, I can't rearrange it to get anything close to y=x-1

Reply 10

username2640523

Original post by dayflower2016

I've been able to get y/(y+1) = x/(x+1) +C
where do I go from here?

How did you get that im confused? Once you seperate into partial fractions and intergrate I think you get natural logs when you intergrate

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Reply 11

dayflower2016

Original post by HFancy1997

How did you get that im confused? Once you seperate into partial fractions and intergrate I think you get natural logs when you intergrate

Posted from TSR Mobile

I did get natural logs, I got lny - ln(y+1) = lnx - ln(x+1) + C
= ln(y/(y+1)) = ln(x/(x+1)) + C
and then exponentiated to get rid of the ln

Reply 12

B_9710

Original post by dayflower2016

I did get natural logs, I got lny - ln(y+1) = lnx - ln(x+1) + C
= ln(y/(y+1)) = ln(x/(x+1)) + C
and then exponentiated to get rid of the ln

Is correct up to that point, now you just need to use boundary conditions.

Reply 13

username2640523

Original post by dayflower2016

I did get natural logs, I got lny - ln(y+1) = lnx - ln(x+1) + C
= ln(y/(y+1)) = ln(x/(x+1)) + C
and then exponentiated to get rid of the ln

If you remove ln the constant would no longer be +C. A new constant would be multiplied so in your case
y/(y+1)=Ax/(x+1)

However let me just check your working out, gimme few minutes :smile:

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Reply 14

username2640523

Do you have what y0=

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Reply 15

username2640523

This is what i got, not what you asked for...Confused :s-smilie:

Reply 16

Zacken

Original post by HFancy1997

This is what i got, not what you asked for...Confused :s-smilie:

Remember that you haven't used the boundary conditions yet.

Reply 17

username2640523

Original post by Zacken

Remember that you haven't used the boundary conditions yet.

Yeah waiting for him to give em but it doesnt look right, we'll see I guess

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Reply 18

Zacken

Original post by HFancy1997

Yeah waiting for him to give em but it doesnt look right, we'll see I guess

Posted from TSR Mobile

Yeah, at a glance, the solution won't be

y = x-1

Reply 19

dayflower2016

I gave all the conditions given to solve the question in the original post :smile:

Thanks guys, I think I've got to grips with these now!

One more thing though, does anybody want to explain to me, in baby steps, how solving non-separable first order differential equations by substitution works? I've watched countless videos, read through the chapters in my textbook, tried the questions and scoured internet forums but it just doesn't make sense to me, everyone seems to do a different thing and I can't translate the method to any question I attempt :'(