@ARosstube You made me worry there ha. It's correct, we have the - before the fraction 1/A(x)B(x)=c/A(x)+d/B(x) ==>1/A(x)B(x)=c/A(x)-d/-B(x) so top and bottom of that fraction is negativised :)
@jungleman68 First notice that 1/(y(y-1))=A/y+B/(y-1) then multiply all by (y(y-1)) to get values for A and B by making them equal zero eg B(y) let y=0 see what A equals. This is known as partial fractions. Then if the top of a fraction [a/b] is the differential of the bottom then the integral is log(b). I hope that has made sense
We are trying to only integrate the x parts wrt x and putting 1/sinx would mean integrating it wrt y which will just make it very complicated in the end. So we are just moving the dx up on the otherside by multiplying and rearranging to have all the x terms on the side with dx and all y terms on side with dy.
At 2:45 I think you switched around the fraction (1-y) to (y-1) without taking the negative to the other side
ARosstube 1 year ago
@ARosstube You made me worry there ha. It's correct, we have the - before the fraction 1/A(x)B(x)=c/A(x)+d/B(x) ==>1/A(x)B(x)=c/A(x)-d/-B(x) so top and bottom of that fraction is negativised :)
burny1 1 year ago
y' = (y^2 + 16)/(t^2 - 25)
atx51288 1 year ago
@atx51288 Is this a correction or a question? Thanks
burny1 1 year ago
how did you integrate 1/(y(y-1))? sorry, I took calc a long time ago. thanks.
jungleman68 1 year ago
@jungleman68 First notice that 1/(y(y-1))=A/y+B/(y-1) then multiply all by (y(y-1)) to get values for A and B by making them equal zero eg B(y) let y=0 see what A equals. This is known as partial fractions. Then if the top of a fraction [a/b] is the differential of the bottom then the integral is log(b). I hope that has made sense
burny1 1 year ago
how do you know to choose sin x as the p(x), why couldnt you do 1/sin x and y(1-y)
relyanddefy 2 years ago
We are trying to only integrate the x parts wrt x and putting 1/sinx would mean integrating it wrt y which will just make it very complicated in the end. So we are just moving the dx up on the otherside by multiplying and rearranging to have all the x terms on the side with dx and all y terms on side with dy.
burny1 2 years ago
ok thanks
relyanddefy 2 years ago