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    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..
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    you havent finished yet! you only found C... u didnt solve the question...
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    (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!
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    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)
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    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!
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    ahh stupid mistake by me...thanks man
 
 
 
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