00:01
In this question, we need to solve the given differential equation using proper substitution.
00:06
So, we have x times dy over dx minus 1 plus x times y is equals to xy square.
00:17
So, first of all, what i will do here, i will divide throughout by divide by y square, what i will get xy raise to minus 2 dy over dx minus 1 plus x raise to y raise to minus 1 is equals to x.
00:39
Now, i will take v to be equals to y inverse, differentiating both side we get dv over dx is equals to minus y raise to minus 2 dy over dx.
00:52
So, let's say this is equation 1, put this value in equation 1, we are going to get y raise to minus 2 dy over dx, it is dv over dx, so it will be minus x dv over dx minus 1 plus x times v because y inverse is v and this is equals to x.
01:13
We can further rearrange this after dividing throughout by minus x.
01:19
So, we get dv over dx plus this will become 1 plus 1 over x times v is equals to minus x, here what we did is divide by minus x, so we got these things...