00:01
Given differential equation x into 3x plus 3y y dash plus y into 9x plus 3y equal to 0.
00:12
Let us divide this by 3 so it is x into x plus y y dash plus y into 3x plus y equal to 0.
00:24
Now let us simplify this equation so divide this by entirely we can take let us take x into let us take x from this expression so it is x square into 1 plus y by x y dash.
00:45
Similarly here also let us take x out it is xy into 3 plus y by x equal to 0.
00:53
So now let us divide this by x so we get x into 1 plus y by x into y dash equal to minus y into 3 plus y by x.
01:10
So let us make a substitution here that is y dash equal to minus y by x into 3 plus y by x by 1 plus y by x.
01:23
So take v equal to y by x then what is dy by dx so y equal to x into v so dy by dx is that is y dash is using the product rule it is v plus x into dv by dx.
01:49
So let us write this in the form v plus x v dash.
01:55
So now let us replace this equation by using v equal to y substituting 2 in 1 our y dash is minus v into 3 plus v by 1 plus v and let us equate this to v plus x v dash.
02:21
So from this expression we can write minus v into 3 plus v by 1 plus v minus v equal to x v dash.
02:37
So v dash is minus v into 3 plus v minus v minus v square by 1 plus v is x v dash by simplifying we get minus 3 v minus v square minus v minus v square by 1 plus v equal to x v dash.
03:02
So we get minus 3 v so minus 2 v square which can be written as minus 2 v square minus 4 v by 1 plus v equal to x v dash.
03:20
So x v dash is minus of taking 2 v common it is 2 v into v plus 2 by v plus 1.
03:33
So we get x v dash is minus 2 v into v plus 2 by v plus 1.
03:41
So now let us separate the variables v dash is nothing but dv by dx.
03:45
So x into dv by dx is minus 2 v into v plus 2 by v plus 1.
03:55
So separating the variables it is dv into v plus 1 by v into v plus 2 into 2 equal to minus integral of dx by dx.
04:10
Now we can integrate this...