00:01
In this problem we are provided with the differential equation x times dy over d x equals to y plus x squared minus 1 the whole squared we are asked to solve this differential equation so let us begin by dividing the entire differential equation by x we have dy over d x to be equal to y over x plus x squared minus 1 the whole squared the whole divided by x next, let us subtract with y over x on both sides.
00:36
We get dy over dx minus y over x which equals to x squared minus 1 the whole squared the whole divided by x.
00:46
So now we can see that this differential equation resembles the bernoulli's form that is dy over dx plus p times y equals to q where p as well as q are functions in terms of x.
01:04
Here we have p to be equal to negative 1 over x and q equals to x squared minus 1 the whole squared the whole divided by x.
01:17
So the general solution of the bernoulli's form is given by y times the integrating factor if which equals to the integral of q times the integrating factor dx, where the integrating factor equals to e raised to the power of integral of p dx.
01:42
So let us begin by finding out the integrating factor.
01:48
So in this case we would have the integrating factor to be equal to integrating factor equals to e raised to the power of p which is negative 1 over x d x.
02:02
The integral of 1 over x is natural log of x.
02:07
So we have e raised to the power of natural log of x.
02:12
This can be rewritten as e raised to the power of natural log of x raised to the power of negative 1 which can further be rewritten as e raised to the power of natural log of 1 over x.
02:27
Since e and natural log are inverse, versus we obtain the answer for the integrating factor to be 1 over x...