Solve the following boundary value problem. ytt = 9yxx, 0 <...

Solve the following boundary value problem.\\ $y_{tt} = 9y_{xx}$, $0 < x < 2$, $t > 0$;\\ y(0,t) = y(2,t) = 0$, $y(x,0) = \frac{1}{8} \sin \pi x$, $y_t(x,0) = 19 \sin 2\pi x$\\ y(x,t) =

Transcript

00:01 All right, so for problem 39, we have to find the solution to this differential equation.

00:06 So when you have a simple differential equation, like when all the coefficients are constants, you can assume that the solution is in the form of e to the power of a constant times the independent variable, which we're going to use x for.

00:20 So we just to derive this, so it's going to be y prime is equal to the k times e to the kx, y double prime is equal to k squared times e to the kx.

00:29 And then we substitute these into our differential equation.

00:33 So it's going to be k squared times e to the kx plus k times e to the kx minus 42 times e to the kx is equal to 0.

00:45 We factor out e to the kx.

00:47 So it's going to be e to the kx is equal to the k squared plus k minus 42 is equal to zero.

00:53 We now find the values for k so that it will make the entire equation equal to zero.

00:59 We know the left part is never going to equal to zero since it's an exponential function, which means we're going to have to rely on the second part to equal to zero.

01:09 And conveniently, it's just a quadratic equation.

01:13 So k squared plus k -mi -4 -2 is equal to zero, and it's factorable.

01:20 So it's k plus 7 times k -minus 6.

01:25 And thus, k is equal to negative 7 and 6.

01:28 And just like that, we found the general.

01:30 Solution to our differential equation, which is y is equal to c1 times e to the negative 7x plus c2 times e to the 6x.

01:40 And now we plug in values to find the two unknown constants.

01:45 So when x is 0, y is to 0.

01:49 So y is 0 is equal to c1, e to the 0 plus c2 times e to the 0.

02:00 E to the 0 is just 1...

Question

Solve the following boundary value problem.\\ $y_{tt} = 9y_{xx}$, $0 < x < 2$, $t > 0$;\\ y(0,t) = y(2,t) = 0$, $y(x,0) = \frac{1}{8} \sin \pi x$, $y_t(x,0) = 19 \sin 2\pi x$\\ y(x,t) =

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