00:01
In this question we have been given an initial value problem, y -daz is equals to 2y square plus xy square, okay, and y -0 is given to be equal to 1 and we need to also check where it attains the minimum value, correct? so that is what we need to check here.
00:23
So let's try to see how can we solve this? so first of all, i will rewrite this equation as d .y divided by dx is equals to x plus two, you can take common here.
00:38
Y square if you take common, you'll get x plus two.
00:41
Okay.
00:42
So now what i can do here is the following things.
00:48
I will just rearrange.
00:49
So dy over y square, and then i will try to integrate.
00:55
And here i will get x plus two.
00:57
Times of dx.
01:00
So integrating this will be minus 1 over y is equals to x squared divided by 2 plus 2x plus constant of integration c1 correct.
01:11
So from this i will get my y of x that is y is equals to 2 over 2c1 plus x squared plus 4x so this is my value of y of x correct so now what else i can do here? so this 2c1 i can consider some other constant.
01:37
So this will implies that y of x i can write as 2 over let's say it is c2 plus x squared plus 4x.
01:48
Correct? this is what i am going to get and minus is obviously there.
01:54
So we need to find out the value of c2 now using the initial condition since it is given that at x is equal to 0.
02:02
Correct.
02:03
So if you put here 1, so it will be minus 2 divided by c2...