Basic math problem with Integrating Factors: Differential Equations

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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.




calculus differential-equations derivatives integrating-factor




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  • 1




    Product rule.$$
    – Clayton
    Aug 9 at 15:16










  • 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 $(xy)'$.
    – user170231
    Aug 9 at 15:19











  • Thanks guys. I get it now.
    – David
    Aug 9 at 16:03














up vote
1
down vote

favorite












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.




calculus differential-equations derivatives integrating-factor




share|cite|improve this question
















  • 1




    Product rule.$$
    – Clayton
    Aug 9 at 15:16










  • 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 $(xy)'$.
    – user170231
    Aug 9 at 15:19











  • Thanks guys. I get it now.
    – David
    Aug 9 at 16:03












up vote
1
down vote

favorite









up vote
1
down vote

favorite











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.




calculus differential-equations derivatives integrating-factor




share|cite|improve this question












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.





calculus differential-equations derivatives integrating-factor





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 9 at 15:13









David

63




63







  • 1




    Product rule.$$
    – Clayton
    Aug 9 at 15:16










  • 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 $(xy)'$.
    – user170231
    Aug 9 at 15:19











  • Thanks guys. I get it now.
    – David
    Aug 9 at 16:03












  • 1




    Product rule.$$
    – Clayton
    Aug 9 at 15:16










  • 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 $(xy)'$.
    – user170231
    Aug 9 at 15:19











  • Thanks guys. I get it now.
    – David
    Aug 9 at 16:03







1




1




Product rule.$$
– Clayton
Aug 9 at 15:16




Product rule.$$
– Clayton
Aug 9 at 15:16












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 $(xy)'$.
– user170231
Aug 9 at 15:19





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 $(xy)'$.
– user170231
Aug 9 at 15:19













Thanks guys. I get it now.
– David
Aug 9 at 16:03




Thanks guys. I get it now.
– David
Aug 9 at 16:03










2 Answers
2






active

oldest

votes

















up vote
1
down vote













Suppose you have



$$xy'+1y=0$$



Notice that we have



this is that we have $$fracddx(xy)=xfracdydx+fracdxdxy=xy'+y$$



by product rule.



Similarly, for $$fracdydx+frac2x1+x^2y=0$$



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



$$(1+x^2) fracdydx+2xy=0$$



which can be written as



$$(1+x^2) fracdydx+fracd(1+x^2)dxcdot y$$



or



$$fracddx(y(1+x^2))=0$$






share|cite|improve this answer




















  • Thank you so much. This helped.
    – David
    Aug 9 at 16:02

















up vote
0
down vote













For your first question you just apply the product rule:



$fracddx(xy)=xfracdydx+y=xy'+x$






share|cite|improve this answer




















  • Thanks a lot. This helped. I get it now.
    – David
    Aug 9 at 16:02










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Suppose you have



$$xy'+1y=0$$



Notice that we have



this is that we have $$fracddx(xy)=xfracdydx+fracdxdxy=xy'+y$$



by product rule.



Similarly, for $$fracdydx+frac2x1+x^2y=0$$



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



$$(1+x^2) fracdydx+2xy=0$$



which can be written as



$$(1+x^2) fracdydx+fracd(1+x^2)dxcdot y$$



or



$$fracddx(y(1+x^2))=0$$






share|cite|improve this answer




















  • Thank you so much. This helped.
    – David
    Aug 9 at 16:02














up vote
1
down vote













Suppose you have



$$xy'+1y=0$$



Notice that we have



this is that we have $$fracddx(xy)=xfracdydx+fracdxdxy=xy'+y$$



by product rule.



Similarly, for $$fracdydx+frac2x1+x^2y=0$$



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



$$(1+x^2) fracdydx+2xy=0$$



which can be written as



$$(1+x^2) fracdydx+fracd(1+x^2)dxcdot y$$



or



$$fracddx(y(1+x^2))=0$$






share|cite|improve this answer




















  • Thank you so much. This helped.
    – David
    Aug 9 at 16:02












up vote
1
down vote










up vote
1
down vote









Suppose you have



$$xy'+1y=0$$



Notice that we have



this is that we have $$fracddx(xy)=xfracdydx+fracdxdxy=xy'+y$$



by product rule.



Similarly, for $$fracdydx+frac2x1+x^2y=0$$



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



$$(1+x^2) fracdydx+2xy=0$$



which can be written as



$$(1+x^2) fracdydx+fracd(1+x^2)dxcdot y$$



or



$$fracddx(y(1+x^2))=0$$






share|cite|improve this answer












Suppose you have



$$xy'+1y=0$$



Notice that we have



this is that we have $$fracddx(xy)=xfracdydx+fracdxdxy=xy'+y$$



by product rule.



Similarly, for $$fracdydx+frac2x1+x^2y=0$$



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



$$(1+x^2) fracdydx+2xy=0$$



which can be written as



$$(1+x^2) fracdydx+fracd(1+x^2)dxcdot y$$



or



$$fracddx(y(1+x^2))=0$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 9 at 15:19









Siong Thye Goh

78.4k134997




78.4k134997











  • Thank you so much. This helped.
    – David
    Aug 9 at 16:02
















  • Thank you so much. This helped.
    – David
    Aug 9 at 16:02















Thank you so much. This helped.
– David
Aug 9 at 16:02




Thank you so much. This helped.
– David
Aug 9 at 16:02










up vote
0
down vote













For your first question you just apply the product rule:



$fracddx(xy)=xfracdydx+y=xy'+x$






share|cite|improve this answer




















  • Thanks a lot. This helped. I get it now.
    – David
    Aug 9 at 16:02














up vote
0
down vote













For your first question you just apply the product rule:



$fracddx(xy)=xfracdydx+y=xy'+x$






share|cite|improve this answer




















  • Thanks a lot. This helped. I get it now.
    – David
    Aug 9 at 16:02












up vote
0
down vote










up vote
0
down vote









For your first question you just apply the product rule:



$fracddx(xy)=xfracdydx+y=xy'+x$






share|cite|improve this answer












For your first question you just apply the product rule:



$fracddx(xy)=xfracdydx+y=xy'+x$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 9 at 15:20









Bruce

31612




31612











  • Thanks a lot. This helped. I get it now.
    – David
    Aug 9 at 16:02
















  • Thanks a lot. This helped. I get it now.
    – David
    Aug 9 at 16:02















Thanks a lot. This helped. I get it now.
– David
Aug 9 at 16:02




Thanks a lot. This helped. I get it now.
– David
Aug 9 at 16:02












 

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