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I was watching a youtube tutorial on Integrating Factors and I'm lost in a part where the derivative of: xy' + 1y becomes xy.

Please I need some clarification on that part.

Secondly.

I was watching a another youtube tutorial on Integrating Factors and I also got lost in this part.

https://i.stack.imgur.com/pmcP6.png

I don't understand how the first line converts to the other, Please I also need some clarification here.

Thanks.

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    $\begingroup$ Product rule.${}$ $\endgroup$ – Clayton Aug 9 '18 at 15:16
  • $\begingroup$ It's not that the "derivative of $xy'+y$ becomes $xy$", but the other way around. By the product rule, the derivative of $xy$ is $xy'+y$, which allows you to write one side of the ODE as $(xy)'$. $\endgroup$ – user170231 Aug 9 '18 at 15:19
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Suppose you have

$$xy'+1y=0$$

Notice that we have

this is that we have $$\frac{d}{dx}(xy)=x\frac{dy}{dx}+\frac{dx}{dx}y=xy'+y$$

by product rule.

Similarly, for $$\frac{dy}{dx}+\frac{2x}{1+x^2}y=0$$

By multiplying integrating factor of $(1+x^2)$, we have

$$(1+x^2) \frac{dy}{dx}+2xy=0$$

which can be written as

$$(1+x^2) \frac{dy}{dx}+\frac{d(1+x^2)}{dx}\cdot y$$

or

$$\frac{d}{dx}(y(1+x^2))=0$$

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  • $\begingroup$ Thank you so much. This helped. $\endgroup$ – David Aug 9 '18 at 16:02
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For your first question you just apply the product rule:

$\frac{d}{dx}(xy)=x\frac{dy}{dx}+y=xy'+x$

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  • $\begingroup$ Thanks a lot. This helped. I get it now. $\endgroup$ – David Aug 9 '18 at 16:02

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