00:01
In this question, we are asked to solve the given differential equation.
00:04
To do that, we'll make a substitution u equals y over x.
00:09
Then, let's calculate du over dx.
00:13
By the quotient rule, we'll get dy over dx multiplied by x minus y multiplied by dx over dx which is 1 divided by x squared.
00:27
Now we need to solve this equation for dy over dx.
00:30
We'll get x squared times du over dx equals dy over dx times x minus y.
00:43
Then, x squared times du over dx plus y equals x times dy over dx.
00:54
Finally, after dividing everything by x, we'll get x times du over dx plus y over x equals dy over dx.
01:07
Then, recall that u equals y over x.
01:10
So, we can replace this term here by u.
01:13
And we'll get that dy over dx equals x times du over dx plus u.
01:25
Then, plug in that expression in the differential equation for dy over dx.
01:32
We'll get 2 multiplied by x du over dx plus u minus u because y over x gets replaced by u.
01:46
And y squared over x squared gets replaced by u squared.
01:50
So, we'll get minus u equals u squared.
01:54
Now, let's try to solve this differential equation.
02:00
We'll get 2x du over dx plus 2u minus u equals u squared.
02:10
Then, we'll get this simplifies to 2x du over dx equals u squared minus u.
02:19
And this is a separable differential equation.
02:22
We can rewrite this as du divided by u squared minus u equals dx divided by 2x...