00:01
For this problem, we're given a differential equation here, and we want to confirm that this y of x is a solution to this differential equation.
00:11
So to do that, we want to take the derivative and second derivative of y of x here to plug into our differential equation.
00:22
So actually, let's rewrite y of x equal to.
00:31
5 halves x to the third minus 3 halves x so that we get y prime of x is equal to 15 halves x squared minus 3 halves and y double prime of x is equal to 15x and now we take these expressions and we plug them in to our differential equation here.
01:09
So let's do that in blue.
01:12
So starting with 1 minus x squared times y double prime is 15x minus 2x y prime is 15 over 2 x squared minus three halves and then we're going to add that to in to times n plus 1 times y of x, which is 5 halves x to the third minus three halves x.
01:50
And then we can go ahead and simplify this.
01:54
This is equal to distributing this first part.
02:00
We get 15x minus 15x to the third minus 15 x to the third minus 15x to the third plus 3x plus.
02:20
And in this case, we are doing the case when capital n is equal to 3.
02:28
That's the case of the lejeundre equation that we're using...