Solve x(dy/dx)=y+x^3

when y=3 and x =1

http://quicklatex.com/cache3/ql_5d8a...a21a68f_l3.png

have i done this correct ?(Thinking)

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- Aug 16th 2011, 09:42 AMdecoy808integrating factor Solve x(dy/dx)=y+x^3
Solve x(dy/dx)=y+x^3

when y=3 and x =1

http://quicklatex.com/cache3/ql_5d8a...a21a68f_l3.png

have i done this correct ?(Thinking) - Aug 16th 2011, 01:56 PMpickslidesRe: integrating factor Solve x(dy/dx)=y+x^3
It all looks good until you say $\displaystyle \displaystyle \int P ~dx = -\frac{1}{x} = \ln x$ which makes little sense.

You are trying to find an integrating factor $\displaystyle \displaystyle I = e^{\int -\frac{1}{x}~dx}$ go from there. - Aug 16th 2011, 03:11 PMSammySRe: integrating factor Solve x(dy/dx)=y+x^3
As

**pickslides**said, you were good until the middle term of this compound equation: $\displaystyle \displaystyle \int P ~dx = -\frac{1}{x} = \ln x\,.$

It should have been: $\displaystyle \displaystyle \int P ~dx = \int-\frac{1}{x} \,dx= -\ln x\,.$ We ignore the integration constant when finding an integrating factor.

The integrating factor is then: $\displaystyle e^{-ln(x)}\,,$ which can be simplified.