00:03
In this question we are given a differential equation dy by dx is equal to 1 upon x cube y where the initial value condition is y of 1 is equal to 2.
00:17
Now we can solve this equation by using the method of separation of variables.
00:28
So now we will separate the variable of x and the variables of y.
00:34
We can see that here it is dy and after doing the cross product we are getting x cube y into dy is equal to dx.
00:45
Now we will be getting y into dy is equal to dx upon take this x cube to the right hand side and it will be in denominator.
00:55
So now after integrating on both sides, integrating both sides we will be getting y into dy will be y square by 2 is equal to integration of x to the power 3 with respect to this.
01:14
So this is basically integration of x to the power minus 3.
01:17
So x to the power minus 3 plus 1 upon minus 3 plus 1 plus constant of integration.
01:24
So that will be equal to y square by 2 is equal to x to the power minus 2 upon minus 2 plus c.
01:34
Similarly we can write it as y square by 2 is equal to minus 1 by 2 upon x to the power 2 plus c.
01:45
Alright.
01:46
Now we have to put our initial value condition that is given by y of 1 is equal to 2.
01:53
Okay now we will use it.
01:54
So the value of y is given to be 2.
01:57
So 2 square upon 2 this is equal to minus 1 upon 2 into 1 square which is 1 plus...