Asked by Betty
integrate dydx=(4sqrt(y)lnx)/x.
Check my answer? I got y=((2ln^2(x)+c)/2)^2. Is that right?
Check my answer? I got y=((2ln^2(x)+c)/2)^2. Is that right?
Answers
Answered by
bobpursley
Take your answer, and take the derivative.
y=((2ln^2(x)+c)/2)^2
y'=2(sqrty)(4lnx/2)
Yep, looks right
y=((2ln^2(x)+c)/2)^2
y'=2(sqrty)(4lnx/2)
Yep, looks right
Answered by
Damon
I assume
dy/dx=(4sqrt(y)ln x)/x
y^-.5 dy = 4 (ln x /x)dx
y^.5 = 2 [ .5 ln^2 x ]
y^.5 = ln^2 x
y = ln^4 x + c
dy/dx=(4sqrt(y)ln x)/x
y^-.5 dy = 4 (ln x /x)dx
y^.5 = 2 [ .5 ln^2 x ]
y^.5 = ln^2 x
y = ln^4 x + c
Answered by
Damon
Yes we agree but I simplified
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