Understanding Integrating Factors and Differential Operators

[Tom McCurdy]

I am going to be gone all day tomorrow at a conference track meet and am unable to ask my teacher how to do integrating factors and differential operators. I leave tomorrow at 9:15 am and was hoping to have some examples to take with me to study.

If someone could help me walk through a these it would be great

integrating factor
$$(x^4+2y)dx-xdy=0$$ answer: $$2y=x^4+cx^2$$
I got it to
$$\frac{dy}{dx}-\frac{2y}{x}=x^3$$

$$e^\int{\frac{-2}{x}}=x^{-2}$$ = IF ... but what do you do then...

$$x^{-2}\frac{dy}{dx}-x^{-2}\frac{2y}{x}=x$$

$$x^{-2}\frac{dy}{dx}-\frac{2y}{x^3}=x$$

so then you get yx^-2 but i don't know what to do from here



more to come

 

[maverick280857]

you get finally

$$\frac{d}{dx}{(yx^{-2})} = x$$

Integrating both sides with respect to x gives,

$$\int d(yx^{-2}) = x^{2} + C$$ **

which gives

[tex]yx^{-2} = x^{2} + C[/itex]<br /> <br /> Multiply both sides by $x^2$ and you're through.<br /> <br /> In step ** you have a total differential under the integral sign which is never a problem to integrate. If you have any more queries, please feel free to ask.<br /> <br /> Cheers<br /> Vivek[/tex]

 

[Tom McCurdy]

Thanks for the help... that's really all i needed to know... was that last step

 

[whozum]

Maybe I'm missing something here but on step ** did you not integrate x incorrectly?

 

[HallsofIvy]

yeah! $$\int x dx= \frac{1}{2}x^2+ C$$!
so $$yx^{-2}= \frac{1}{2}x^2+ C$$.

 

[maverick280857]

Hi everyone

Sorry for that mistake...must've forgotten to use my brain then ;-)

Cheers
Vivek