1. ## Homog Lin... #3

3.) Solve the following Dif EQ, given the indicated initial conditions:

$\frac{d^4y}{dt^4} - y = 0$, $y(0) = y'(0) = y''(0) = 0, y'''(0) = 1$

Work:

So m^4 - 3m^3 + 3m^2 - m = 0

m*(m-1)*(m^2 + m + 1)

Thus, m = 0 or 1.

So, the general sol'n would be:

$y = c_1 + c_2e^{t}$

So, with the conditions:

0 = c_1 + c_2e^(0)

So c_1 = -c_2

Then, would I find the 3rd derivative? Not really sure.

2. Originally Posted by Ideasman
3.) Solve the following Dif EQ, given the indicated initial conditions:

$\frac{d^4y}{dt^4} - y = 0$, $y(0) = y'(0) = y''(0) = 0, y'''(0) = 1$

Work:

So m^4 - 3m^3 + 3m^2 - m = 0

m*(m-1)*(m^2 + m + 1)

Thus, m = 0 or 1.

So, the general sol'n would be:

$y = c_1 + c_2e^{t}$

So, with the conditions:

0 = c_1 + c_2e^(0)

So c_1 = -c_2

Then, would I find the 3rd derivative? Not really sure.
Hmm, I am definitely going without much sleep. Sorry, it should be m^4 - m = 0. But, it turns out it's the same exact answer. That is, factor:

m*(m-1)*(m^2 + m + 1)...

etc etc, just not sure how to proceed after.

3. 3.) Solve the following Dif EQ, given the indicated initial conditions:

$\frac{d^4y}{dt^4} - y = 0$, $y(0) = y'(0) = y''(0) = 0, y'''(0) = 1$

Okay, my second attempt.. I def. messed up the first time.

The auxiliary equation is m^4 - 1 = 0.

We get: $(m - 1)*(m + 1)*(m^2 + 1) = 0$

Thus, $m = 1, -1 , i, -i$

So, $y = c_1e^t + c_2e^(-t) + c_3e^t(sin(x) + c_4e^tcos(x)$

Is this right? I think it is. So, if this is right, then I can find [tex]y', y'', y''' and solve the equations. However, wouldn't it be hard trying to solve the equations for $c_1, c_2, c_3, c_4$... it just feels as there is too much unknown..

4. Originally Posted by Ideasman
3.) Solve the following Dif EQ, given the indicated initial conditions:

$\frac{d^4y}{dt^4} - y = 0$, $y(0) = y'(0) = y''(0) = 0, y'''(0) = 1$

Okay, my second attempt.. I def. messed up the first time.

The auxiliary equation is m^4 - 1 = 0.

We get: $(m - 1)*(m + 1)*(m^2 + 1) = 0$

Thus, $m = 1, -1 , i, -i$

So, $y = c_1e^t + c_2e^(-t) + c_3e^t(sin(x) + c_4e^tcos(x)$

Is this right? I think it is. So, if this is right, then I can find [tex]y', y'', y''' and solve the equations. However, wouldn't it be hard trying to solve the equations for $c_1, c_2, c_3, c_4$... it just feels as there is too much unknown..
Aside from a wild set of parenthesis and a LaTeX coding error, this is a valid way to express the solution. You will have to solve some simultaneous equations to get the values of the constants, so you still have some work ahead of you.

-Dan