Understanding Integrating Factors and Differential Operators

AI Thread Summary
The discussion focuses on finding integrating factors and solving a differential equation involving the expression (x^4 + 2y)dx - xdy = 0. The user successfully derived the equation to the form dy/dx - (2y/x) = x^3 and identified the integrating factor as x^{-2}. However, confusion arose regarding the integration step, where the correct integral of x was initially miscalculated. After clarification, it was confirmed that integrating both sides leads to the correct solution, yx^{-2} = (1/2)x^2 + C, demonstrating the importance of careful integration in solving differential equations. The conversation highlights the collaborative nature of learning and problem-solving in mathematics.
Tom McCurdy
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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
 
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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

yx^{-2} = x^{2} + C[/itex]<br /> <br /> Multiply both sides by x^2 and you&#039;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
 
Thanks for the help... that's really all i needed to know... was that last step
 
Maybe I'm missing something here but on step ** did you not integrate x incorrectly?
 
yeah! \int x dx= \frac{1}{2}x^2+ C!
so yx^{-2}= \frac{1}{2}x^2+ C.
 
Hi everyone

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

Cheers
Vivek
 
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