There actually isn't all that much to this section. All we are going to do now is work a quick illustration using Laplace transforms for a 3rd order differential equation therefore we can say that we worked, at least one problem for a differential equation whose order was larger than 2.
Everything which we know from the Laplace Transforms section is even valid. The only new bit which we'll require here is the Laplace transform of the third derivative. We can find this from the general formula that we gave while we first started looking at solving initial value problems with Laplace transforms. Now there is that formula,
L ({y′′′} + s3Y(s) - s2 y(0) - sy′ (0)- y′′(0)
Now there the example for this section
Example: Solve the following initial value problem.
y′′′ - 4y'' = 4t + 3u6(t) e30 - 5t; y(0) = -3; y''(0) = 4
As always we first require making sure the function multiplied with the Heaviside function has been correctly shifted.
y′′′ - 4y'' = 4t + 3u6(t) e - 5(t- 6)
This has been properly shifted and we can notice that we're shifting e-5t. All we require to do now is take the Laplace transform of everything, plug into the initial conditions and answer for Y (s).
Doing all of this provides,
s3Y(s) - s2y(0) - sy'(0) - y''(0) - 4(s2Y(s) - sy(0) - y'(0)) = (4/s2) + ((3 e-6s)/(s + 5))
(s3 - 4s2)Y(s) + 3s2 - 13s = (4/s2) + ((3 e-6s)/(s + 5))
(s3 - 4s2)Y(s) = (4/s2)- 3s2 + 13s + ((3 e-6s)/(s + 5))
Y(s) = (4- 3s4 + 13s3)/(s4(s - 4))+ ((3 e-6s)/ ((s4(s - 4)) (s + 5)))
Y (s) = F(s) + 3e-6 G(s)
Now we need to partial fraction and inverse transform F(s) and G(s). We'll leave it to you to verify the details.
Now we require to partial fraction and inverse transform F(s) and G(s). We'll leave this to you for verify the details.
F(s) = (4- 3s4 + 13s3)/(s4(s - 4)) = ((17/64)/(s - 4)) - (209/s) - ((1/16)/s2) - ((1/4)(2!/2!)/s3) - ((1(3!/3!)/s4)
f(t) = (17/64)e4t - (209/64) - (1/16) t - (1/8)t2 - (1/6)t3
G(s) = 1/(s2(s- 4)(s + 5)) = ((1/44)/(s - 4)) - ((1/225)/(s + 5)) - ((1/400)/s) - ((1/20)/s2)
g(t) = (1/144)e4t - (1/225)e-5t - (1/400) - (1/20)t
Okay, we can here find the solution to the differential equation. Starting along with the transform we find,
Y (s) = F(s) + 3e-6 G(s) ⇒ y(t) = f(t) + 3u6(t) g(t -6)
Here f(t) and g(t) are the functions demonstrated above.
Okay, here is the one Laplace transform illustration with a differential equation with order greater than 2. As you can notice the work in the same except for the fact that the partial fraction works (that we didn't demonstrate here) is liable to be messier and more complex.