3 let px and qx be given functions and define lux ux pxux...

Name: 3 let px and qx be given functions and define lux ux pxux qxux for every function ux that is twice differentiable show that if ux and vx are twice differentiable functions and a in mathbbr t 82619
Uploaded: 2020-09-02T03:06:24-08:00
Duration: 7 min 16 s
Channel: Victoria Dollar
Description: 3 let px and qx be given functions and define lux ux pxux qxux for every function ux that is twice differentiable show that if ux and vx are twice differentiable functions and a in mathbbr t 82619

3. Let $p(x)$ and $q(x)$ be given functions, and define $L(u(x)) = u''(x) + p(x)u'(x) + q(x)u(x)$ for every function $u(x)$ that is twice differentiable. Show that if $u(x)$ and $v(x)$ are twice differentiable functions, and $a \in \mathbb{R}$, then $L(u(x) + av(x)) = L(u(x)) + aL(v(x))$.

Transcript

00:01 So for this problem, we are asked for part a that we're going to use the equation y equals 1 over mu times the integral of mu, q of x, dx.

00:15 And so what i'm going to do first is i'm going to multiply mu to both sides.

00:20 So mu y equals the integral of mu, q of x, dx.

00:27 And then what i want to do is i want to find d over dx for both sides.

00:33 We're essentially just going to be solving these.

00:35 Backwards.

00:38 And so now what i have is i have my integrating factor and my variable.

00:43 So what this tells me is i have mu, d, y over dx plus mu p of x, y, because those are the restrictions to be able to use that.

00:56 And then, of course, mu q of x, because this derivative and the integral cancel.

01:03 So if i were to divide everything by mu, i would have d, y, over d, x.

01:09 Plus p of x y equals q of x which of course is our equation that we're trying to get as our goal now for part b it is asking us to um it is asking us to go ahead and find um the uniqueness of this so in order to do we're actually going to be looking at the initial value problem so what we want to do is i want to go ahead and plug this in and i will want to, so if i, let's say let the integral of mu, q of x, dx, equal capital r of x plus c.

02:05 Or actually, let's go and make this lowercase.

02:10 Then what i could do is i can let y equals 1 over mu, and then instead of this integral, i could do r of x plus c, so which means that if, let's wanted to use the initial condition, which is y if x not equals y not, then what i would do is plug this in.

02:44 So y not equals 1 over mu.

02:49 Oops, r of x not plus c.

02:58 So then what i want to do is that i want to go ahead and solve for c.

03:06 Okay, so what i want to do is i want to multiply it by mu.

03:09 So i'm going to have mu y of zero equals r of x plus uh, uh, not plus c.

03:16 So then c is going to equal mu of y not minus r of x not...