1 a form the partial differential equation of the following...

1. (a) Form the partial differential equation of the following (i) $z = \frac{x^2}{a^2} + \frac{y^2}{b^2}$ (ii) $z = f(x + ay) + g(x - ay) (b) Use Lagrange's Method to find the general solution of $\frac{y^2z}{x} \frac{\partial z}{\partial x} + xz \frac{\partial z}{\partial y} = y^2$. (c) Solve the heat conduction equation $\frac{\partial^2 u}{\partial x^2} = \frac{1}{2} \frac{\partial u}{\partial t}$ over $0 < x < 3$, $t>0$ for the boundary conditions $u(0,t) = u(3,t) = 0$ and the initial condition $u(x, 0) = 5 \sin 4\pi x$.

Transcript

00:01 Well, i certainly don't have enough space to do all of the problems, but i'll do a few of them.

00:07 So remember that in general, p is going to be partial z with respect to x, and q is partial z with respect to y.

00:20 So we get the characteristic equations, which are the equations of the characteristics, and they look like x.

00:38 Dot equals a y and z dot equals z so z dot the thing on the left hand side is equal to z dot because it's x p plus y q okay and so the first one of these tells us that x is some let's call it x zero let's call it c1 it's a constant so c1 e to the tau and this one tells us us that y is another constant c2 times e to the tau where tau is going to be our parameter along our characteristics and then this one gives us a z is some c3 e to the tau okay and we can eliminate tau between those in a variety of ways so for instance y over x is a constant i'll call that k1 and let's say z over x is another constant i'll call k2.

01:54 So these constants are arbitrary.

01:59 And if i take any arbitrary function of those two constants and set it equal to zero, that will give me a solution.

02:14 So i take any old function f of the quantity y over x and z over x, any arbitrary function of those two variables will work.

02:38 Okay, if we had some initial conditions, we could maybe figure out what f is, but we're not given that information.

02:48 So in part b, we have, so p is, excuse me, it's z.

02:57 So it's z partial, z with respect to x equals minus x.

03:06 So why does not appear? okay? why does not appear in this equation? so any y dependence is going to be entirely arbitrary.

03:21 But i can treat this as an ode, an ordinary differential equation, and rewrite it like this.

03:30 Zdz plus x, dx, equals zero.

03:35 And then i can integrate it.

03:37 So i get half z squared plus.

03:40 Half x squared is equal to a constant and that constant is an arbitrary function of y now put a one -half in there give rid of those one -haves and so i can get z squared plus x squared is some arbitrary function c1 of y so any pick any function of y and this tells me what z is.

04:23 And that's my solution.

04:25 Again, we don't have any initial conditions or anything to work with.

04:35 Number c, we got tangent of x times p plus tangent of y times q is equal to tangent of z.

04:54 Okay, so now we're write out our characteristic equations, and they tell us that dx, d, tau, is equal to the tangent of x, and dy, d tau is equal to the tangent of y, and dz, d tau is the tangent of z.

05:28 All right, so we can integrate those.

05:35 So the first step is to solve these for the characteristics.

05:40 So if i take the ratio of the first two, so i got x.

05:49 .

05:49 ...

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