Question

Differential Equations Project - must be completed in Maple 2018 program

NEED ALL PARTS OF THE PROJECT (A - F)

In this Maple lab you learn the Maple commands for computing Laplace
transforms and inverse Laplace transforms, and using them to solve initial
value problems.

A. Quite simply, the calling sequence for taking the Laplace transform of
a function f(t) and expressing it as a function of a new variable s is

laplace(f(t),t,s) .

The command for computing the inverse Laplace transform of a function
F(s) and expressing it as a function of a variable t is

invlaplace(F(s),s,t) .

B. Laplace transform techniques are most useful when the right hand side
of the differential equation is either discontinuous or contains the Dirac
delta function. The names in Maple for the functions that arise in these
situations are:

for the unit step function with a step from 0 to 1 at x = 0 :

Heaviside(t)

for the Dirac delta function:

Dirac(t)

For the remainder of the lab we'll go through a number of examples.

1. Load the integral transforms package:

with(inttrans):

C. Let's begin by having Maple compute the Laplace transform of the
f(t) = t^2*sin(pi*t), and then inverting it. The name Pi (note the
capitalization) is reserved in Maple for the constant 3.1415....

f:=t^2*sin(Pi*t);
F:=laplace(f,t,s);
ff:=invlaplace(F,s,t);

D. Now let's solve a straightforward initial value problem using Laplace
transform methods, but having Maple compute the transform and invert
it for us. We'll solve

y'' + y' + 3*y = cos(t), y(0) = 2, y'(0) = -1 .

The example illustrates that we can apply the operator laplace to the
entire o.d.e. at once. The fact that Maple doesn't know what the function
y(t) is doesn't matter; it simply returns the name laplace(y(t),t,s) for
the unknown Laplace transform, meanwhile applying the rules for
turning differentiation into multiplication.

de1:= diff(y(t),t,t) + diff(y(t),t) + 3*y(t) = cos(t);
Lde1:=laplace(de1,t,s);

We could now apply the initial conditions and solve for the Laplace transform
Y(s) of the solution by hand. Instead, we'll use the substitution command subs
and the solve command:

Lde1:=subs(y(0)=2,D(y)(0)=-1,Lde1);
Y1:=solve(Lde1,laplace(y(t),t,s));
y1:=invlaplace(Y1,s,t);

Please do upload all parts... THANKS!

E. Now let's solve an initial value problem with discontinuous right hand side:

y'' + 0.5y' + 4y = f(t) , y(0) = 1, y'(0) = 0

where
  
f(t) = cos(3t) for 0 < t < 10*pi, f(t) = 0 otherwise.

This i.v.p. models a damped forced spring-mass system, for which the periodic
forcing is removed at time 10*pi.

f2:=(1-Heaviside(t-10*Pi))*cos(3*t);
de2:=diff(y(t),t,t) + (1/2)*diff(y(t),t) + 4*y(t) = f2;
Lde2:=laplace(de2,t,s);
Lde2:=subs(y(0)=1,D(y)(0)=0,Lde2);
Y2:=solve(Lde2,laplace(y(t),t,s));
y2:=invlaplace(Y2,s,t);

We can't tell much from the formula for y2(t), so let's look at the graph.

plot(y2,t=0..15*Pi);

It makes sense: there is a transient until the system oscillates in step with the periodic
forcing; the system then oscillates until the forcing is removed, at which point the
motion damps out.

F. Finally, let's consider a system with an impulsive force:

y'' + 4y = g(t) , y(0) = 1, y'(0) =2
where

g(t) = 3*delta(t-4) .

This i.v.p. models an undamped, unforced spring-mass system to which a sharp
blow of magnitude 3 is applied at time t = 4 .   

f3:=3*Dirac(t-4);
de3:=diff(y(t),t,t) + 4*y(t) = f3;
Lde3:=laplace(de3,t,s);
Lde3:=subs(y(0)=1,D(y)(0)=2,Lde3);
Y3:=solve(Lde3,laplace(y(t),t,s));
y3:=invlaplace(Y3,s,t);

Again, we can't tell much from the formula for y3(t), so let's look at the graph.

plot(y3,t=0..10);

The blow is in the direction opposite to the motion at time 4 , and induces a new
oscillation at a smaller amplitude.

Now start your own Maple session, and solve the following three problems using
Laplace transforms. Note that plotting the solutions is involved.

Problem 1.
Find the solution of the following intial value problem, and plot it on the interval
from t = 0 to t = 4*Pi.

y'' + 2y' +2y = f(t), y(0) = 0, y'(0) = 1

where

f(t) = 2 for pi < t < 2pi , and is zero otherwise

Problem 2.
Find the solution of the following intial value problem, and plot it on the interval
from t = 0 to t = 8*Pi.

y'' + 2y' +2y = g(t), y(0) = 0, y'(0) = 0

where g(t) is the cosine function for all time, except for a sharp blow of
magnitude 3, in the positive direction, at time (5/2)pi .
  

Problem 3.
Laplace transform methods work equally well, with no extra effort on your part,
for higher order o.d.e.'s, as the following problem illustrates. Solve the following
initial value problem, and plot the solutiion on the interval from t = 0 to t = 12.
Remark: you can express the second derivative by diff(y(t),t$2) , and similarly for
third and fourth derivatives. You can express the initial conditions involving the
higher derivatives using (D@@3)(y)(0)=2/3 and so on.

y'''' + 3y''' + 6y'' + 12y' + 8y = h(t), y(0) = 1, y'(0) = -2, y''(0) = 1/2, y'''(0) = 2/3

where h(t) corresponds to a sharp blow of magnitude 24, in the negative
direction, at time 2 (but h(t) is zero otherwise).

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