I need help with this problem: Setting y/x=u, solve: xy' = (y-x)^3 + y; y(1) = 3/2 Thanks!
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Originally Posted by ntfabolous I need help with this problem: Setting y/x=u, solve: xy' = (y-x)^3 + y; y(1) = 3/2 Thanks! Using the hint given and assuming u is a function of x. Taking the derivative we get Now subbing this into the equation we get Simplifing and factoring we get Integrating both sides we get From here it is just back subbing good luck.
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