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# diff equation watch

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1. Solve diff eq...

y(dy/dx)= tanx (given that y=1 & x=pi/4)

so..i did...I ydy= I sinx/cosx
y^2/2= ln(secx) + c
substitute values..; 1/2= ln (sec pi/4) + c
1/2= ln(1/cospi/4) + c...
1/2= ln(2/rt2) + c...
c= 1/2 - ln(2/rt2)

wt have i done wrong..
2. you havent finished yet! you only found C... u didnt solve the question...
3. (Original post by Nandoz)
Solve diff eq...

y(dy/dx)= tanx (given that y=1 & x=pi/4)

so..i did...I ydy= I sinx/cosx
y^2/2= ln(secx) + c
substitute values..; 1/2= ln (sec pi/4) + c
1/2= ln(1/cospi/4) + c...
1/2= ln(2/rt2) + c...
c= 1/2 - ln(2/rt2)

wt have i done wrong..
I can't immediately see a problem with that. You can cancel c down to be
c=0.5[1-ln2] if you want to be pedantic, giving
y^2 = 2ln|secx| + 1 - ln2
y = root { ln[0.5(secx)^2] + 1

But what you've done it all correct I think!
4. ye..bt in the end my answer comes out as....y^2/2=lnsecx + (1/2 - ln 2/rt2)
y^2= 2(lnsecx + (1/2 - ln 2rt2)...y= rt [2(ln sec x + (1/2 - 2/rt2)

but the answer says...y= rt(1- ln2 - 2lncosx)
5. y^2/2=lnsecx + (1/2 - ln 2/rt2)
y^2=2 [lnsecx + (1/2 - ln 2/rt2)]
y = sqrt [2lnsecx + 1 - 2 ln 2/rt2]
y = sqrt [ -2lncosx + 1 - ln (2/rt2)²]
y = sqrt [ -2lncosx + 1 - ln (4/2)]
y = sqrt [ -2lncosx + 1 - ln (2)]

i dont know why did u remove the ln from the second step to the third step!
6. ahh stupid mistake by me...thanks man

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Updated: June 13, 2006
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