How to solve the seperable differential equation and when using the following initial condition: y(1)=2 ?

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Answer 1 · 2016-05-09T21:26:30

#y = root(6)(6ln(x)+6x+58)#

#dy/dx = (1+x)/(xy^5)#

#=> y^5dy = (1+x)/xdx=(1/x+1)dx#

#=> inty^5dy = int(1/x+1)dx#

#=> y^6/6 = ln(x)+x + C#

#=> y^6 = 6ln(x)+6x + C#

By the initial condition of #y(1)=2#, we have

#2^6 = 6ln(1)+6+C = 6+C#

#=> C = 2^6 - 6 = 58#

#=> y^6 = 6ln(x)+6x+58#

#:. y = root(6)(6ln(x)+6x+58)#