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# C4 Differential equations question watch

1. Hi I'm having some trouble with this question - I can't see what I've done wrong but I'm not coming out with the same answer as the book. Any help?

Nb: there is the possibility that the answer in the book might be wrong - it's done that before.

Q: solve the differential equation

dy/dx = 2ysec(^2)2x, given that y=2 when x=0

dy/dx = 2ysec(^2)2x
?2y dy = ?sec²2x dx
ln(2y) = 1/2 tan2x+c

@ x=o,y=2

ln(4)=1/2 tan(0) + c
c = ln(4)

=>ln(2y) = 1/2 tan2x+ln(4)

2y = e^(1/2 tan2x + ln(4))
2y = e^1/2 tan2x x e^ln(4)
2y = 4e^1/2 tan2x
y = 2e^1/2 tan2x

y = 2e^tan2x

what happened to the 1/2 ?

Thanks!
2. (Original post by SparksInTheSky)
?2y dy = ?sec²2x dx
ln(2y) = 1/2 tan2x+c
Not sure what that first line is meant to be, but:

and your error is propagated through the rest of your working; and by the looks of things doesn't show up until the end.

PS: Book's right.
3. (Original post by Farhan.Hanif93)
The book right, you're wrong.
Corrected As per ghostwalker's post.
4. (Original post by nuodai)
Corrected As per ghostwalker's post.
Oops, didn't notice that.
5. Yeah should be 1/2ln(2y), or keep the 2 with the x
6. (Original post by ghostwalker)
Not sure what that first line is meant to be, but:

and your error is propagated through the rest of your working; and by the looks of things doesn't show up until the end.

PS: Book's right.
Yeah that was meant to be an intergral sign, dunno why it didn't come out right.

Yeah I see it now, integral 1/2y = 1/2 integral 1/y = 1/2ln(y)

then it all works out. Thanks! (feel a bit stupid now...!)
7. (Original post by SparksInTheSky)
Yeah that was meant to be an intergral sign, dunno why it didn't come out right.
Have a look at using LaTex; there's a link on the forum page, and at the top of this thread.

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