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

• Aug 16th 2011, 10:42 AM
decoy808
integrating 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, 02:56 PM
pickslides
Re: integrating factor Solve x(dy/dx)=y+x^3
It all looks good until you say $\displaystyle \int P ~dx = -\frac{1}{x} = \ln x$ which makes little sense.

You are trying to find an integrating factor $\displaystyle I = e^{\int -\frac{1}{x}~dx}$ go from there.
• Aug 16th 2011, 04:11 PM
SammyS
Re: integrating factor Solve x(dy/dx)=y+x^3
Quote:

Originally Posted by pickslides
It all looks good until you say $\displaystyle \int P ~dx = -\frac{1}{x} = \ln x$ which makes little sense.

You are trying to find an integrating factor $\displaystyle I = e^{\int -\frac{1}{x}~dx}$ go from there.

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

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

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