00:01
Hi, here in this question we are given a differential equation which is 6y square dx minus x times 2x cube plus y dy equals to 0.
00:14
So here on further simplification we have equation dx upon dy equals to x to the power 4 minus 3y square plus x upon 6y.
00:27
So here our equation can be rewritten as here we have dx upon dy minus x upon 6y equals to x to the power 4 upon 3y square.
00:40
Now here by dividing the equation with 1 upon y square here we have dy upon dx multiplied with 3 minus x to the power 4 plus 1 upon 2 times yx cube equals to minus 1 upon y square.
00:59
Now here let this be our equation 2.
01:02
This is our original equation.
01:03
Let this be our equation 1.
01:06
Now let us assume that v equals to 1 upon x cube which implies the value of dv upon dy equals to minus 3 times dx upon dy divided by x to the power 4.
01:19
So substituting this value in the equation 2, here our equation 2 can be rewritten as dv upon dy plus 1 upon 2y multiplied with v equals to minus 1 upon y square.
01:32
Let this be our equation 3.
01:34
So here using integrating factor method we can obtain the solution...