00:01
Alright, so start off this problem.
00:02
Let's go ahead and divide everything by x, y, squared.
00:07
So i'll have d -y -d -x equals to y divided by x minus, let's see, one.
00:19
All right, or sorry, this would be x squared divided by y squared.
00:28
Okay.
00:29
And so now with this, we can go ahead and use our substitution.
00:32
I am going to use a substitution u equals y divided by x and so by that logic i'm going to have y equal x times u or d -y -d -x equal to x d -d -x plus u so let's go ahead and plug these in so i'll have x -d -x plus u equals to x -u divided by or sorry equals to u minus one divided by u squared all right and i can subtract you from both sides so this and this will cancel out and then i'll have x d u d x equals to negative one divided by u squared i can move the x terms to the right hand side and the u terms to the left hand side so i have u squared du is equal to the integral of x d x or sorry one divided by x d x okay so on the left hand side when we integrate we should get negative u cubed divided by three and this would equal to the natural log of the absolute value of x plus c i can multiply everything by three so i'll have negative u cubed equals to three natural log of the absolute value of x plus c and i can go ahead and and plug in my u equals y divided by x.
02:19
So i have y cubed, divided by x cubed, equals to three natural log of x plus c.
02:26
I'll multiply by the negative...