00:01
I have to solve this, the national recognition is given to this.
00:04
All the different equation.
00:05
So what we can do here, you can put like u equal is y negative x and y negative step x here.
00:12
You equals y negative x.
00:14
That's what we go.
00:14
We find out d u over dx that equals d y over d x negative 1.
00:21
Let's be called here from here we get d y over d x.
00:23
It will be 1 plus d u over d x means we're just bringing out y and placing it as u in terms of you.
00:30
So this is coming up to be 1 negative ux negative u squared x u.
00:37
So we get d u x equals negative x u negative x u squared.
00:46
Next we divide by u square both sides we get 1 over u squared d u positive x times 1 over u equals negative x2.
00:59
So from here if we solve this coming up to be negative 1 over u squared du by d f plus x negative 1 over u equal x q kb a ven x now what we can do is set here y equals 1 over u if is that now y equals 1 over u we have d y over d x that's coming out to be d capital y over d x negative 1 over u squared d u by d so we just bring it out in the linear form that's come out to be d y over d x negative x to negative x y equal xx non -premaud.
01:44
Why not an integrating factor that's coming up to be indicating factor equals e to the power we have a integrated negative x d x that will be e to the power negative x square over 2 so we get y times a to the power negative x square over 2 in the vector.
02:01
So we're not going to be an equation equals t plus we have cube that is x cube and times 8 to the power negative xpure over 2 dx.
02:13
So we divide it now.
02:15
Using here integration by parts, so the first function...