SOLUTION: Solve the initial value problem y'+y/2x=xy^-3,y(1)=2.Please help me.

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Question 976251: Solve the initial value problem y'+y/2x=xy^-3,y(1)=2.Please help me.
Answer by t0hierry(194) About Me  (Show Source):
You can put this solution on YOUR website!
y^3dy + y^4/2xdx = x
u = y/sqrt(x)
y = u sqrt(x)
y' = u' sqrt(x) + u/2sqrt(x) = x/y^3
u'x + u/2 = sqrt(x)^3/y^3 = 1/u^3
xdu = (1/u^3 - u/2)dx
du/(1/u^3 - u/2) = dx/x
du 2u^3/(2-u^4) = dx/x
call v = 2 - u^4
then dv = -4 u^3 du
-2 dv/v = dx/x
integrate
v^(-2) = 1/x + constant
y(1) = 2 means 1 + constant = 2 - 2^4/1 = 2 - 16 = -14
constant = 15
If I didn't make any mistake (please check) then
(2 - y^4/x^2)^(-2) = 1/x + 15