00:01
In this question we are asked to solve this differential equation.
00:03
So let's start to solve this problem.
00:07
The given differential equation is dy over d x is equal to y x into y square plus 2.
00:15
We can write the given differential equation as dy over d x is equal to x into y cube plus 2y.
00:26
From this we obtained dy over d x minus 2y is equal to x into y cube.
00:33
Divide this whole equation by y cube 1 over y cube, dy over d x minus 2 over y square is equal to x.
00:44
Now we substitute 1 over y square is equal to v.
00:49
After differentiating this with respect to x, we obtained negative 2 over y cube dy over d x is equal to d b over d x.
01:04
1 over y cube, dy over dx is equal to negative 1 over 2 bv over d x after these two substitution the given differential equation becomes negative 1 over 2 dv over dx minus 2v is equal to x dv over dx plus 4v is equal to negative 2x here this differential equation present in the form of linear differential equations because standard form of linear differential equation is given by dv over d x plus px into v is equal to qx.
01:53
By comparing these two we obtained px is equal to 4 and q x is equal to negative 2x...