### Author Topic: Question from Monday's Lecture  (Read 866 times)

#### Tristan Fraser

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##### Question from Monday's Lecture
« on: January 16, 2018, 10:04:18 PM »
When we were (re)introduced to the following Cauchy problem:
(i) $$u_{tt} - u_{xx} = 0$$
(ii) $$u|_{t =\frac{x^2}{2}} = x^3$$
(iii) $$u_{t} |_{t = \frac{x^2}{2}} = 2x$$

And we took the general solution of

(from i: 1) $$u(x,t) = \phi(x +t) + \psi (x-t)$$
and applying the conditions (ii,iii)
(from ii: 2) $$u(x, \frac{x^2}{2}) = \phi(x + \frac{x^2}{2}) + \psi(x - \frac{x^2}{2}) = x^3$$
(from iii : 3) $$u_{t} (x,\frac{x^2}{2}) = \phi'(x + \frac{x^2}{2}) - \psi'(x - \frac{x^2}{2}) = 2x$$
(2') $$3x^2 = (1+x) \phi'(x + \frac{x^2}{2}) + (1-x)\psi'(x - \frac{x^2}{2})$$

But from there we wrote that :
$$\phi'(x+ \frac{x^2}{2}) = \frac{\begin{vmatrix} 2x & -1 \\ 3x^2 & 1-x \end{vmatrix}}{\begin{vmatrix} 1 & -1 \\ 1+x & 1-x \end{vmatrix}} = \frac{2x+ x^2}{2} = x+ \frac{x^2}{2}$$

and similarly for $$\psi'(x - \frac{x^2}{2}) = -x + \frac{x^2}{2}$$

Where did these determinants come from?

#### Ioana Nedelcu

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##### Re: Question from Monday's Lecture
« Reply #1 on: January 16, 2018, 11:25:05 PM »
Cramer's rule for solving systems of equations (should be familiar if you took MAT224)

I solved the equations using substitution/ elimination (read: the usual way) and got the same thing but with more steps
« Last Edit: January 17, 2018, 12:57:43 AM by Victor Ivrii »