00:01
Derivative of y with respect to x is equal to xy square minus cos x sin x divided by y 1 minus x square and y at 0 is equal to 2.
00:21
This is the boundary condition.
00:23
So, what we can do is we can cross multiply and we can write it as xy square minus cos x sin x dx minus y 1 minus x square dy is equal to 0.
00:42
So, we can see that the equation is of form mdx plus ndy equals to 0.
00:58
So, when we compare these two equations, we get m is equal to xy square minus cos x sin x and n is equal to minus of this term that is y x square minus 1.
01:20
So, we will now partially differentiate m with respect to y and we get 2xy and when we differentiate n partially differentiate n with respect to x we get 2xy.
01:35
So, since del m divided by del y is equal to del n divided by del x.
01:46
So, we can say that equation is exact.
01:53
Now we can write the solution...