Transcript Maple Intro and assignment 2

Eng. 6002 Ship Structures 1
Hull Girder Response Analysis
Introduction to Maple
Overview
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PURPOSE: To illustrate the use of dsolve to find
exact solutions of many differential equations or
systems of equations, with or without initial
conditions.
NOTE 1. Maple has a powerful symbolic
differential equation solver called dsolve. This
command directs Maple to seek an exact
symbolic expression for the solution of a given
differential equation, or a system of differential
equations, with or without initial conditions.
NOTE 2. Further information about dsolve can be
obtained by issuing the command ‘?dsolve’.
Overview cont.
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IMPORTANT NOTE: dsolve insists that
coefficients be entered as rational, rather
than floating point, numbers. For example,
you should enter a coefficient as 1/2 or
1/4, rather than 0.5 or 0.2.
Example 1: A 1st order linear equation
Find the general solution of the differential
equation dy/dt = -2 + sin(t) - (1/4)y.
 Also find the solution that satisfies the
initial condition y(0) = 2.
 SOLUTION. The first step is to enter the
differential equation.
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eq1 := diff(y(t),t) = -2 + sin(t) - 1/4*y(t);
Example 1: A 1st order linear equation
Now dsolve can be used to find the
general solution of this equation.
 In its simplest form the command has two
arguments;
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the first is the equation to be solved
the second is the variable to be solved for.
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dsolve(eq1,y(t));
Example 1: A 1st order linear equation
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This is the general solution of the given
differential equation. Note that Maple writes the
arbitrary constant as _C1, and that in this case
the constant follows the term exp(-t/4) that it
multiplies.
To find the solution that also satisfies the initial
condition, we include the initial condition in the
dsolve command, using braces to delimit the
problem to be solved. Also we assign the name
sol to the result.
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sol := dsolve({eq1,y(0)=2},y(t));
Example 1: A 1st order linear equation
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We can now assign a name, if we wish, to
the expression for y(t) on the right side of
the last result, as follows:
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> y1 := rhs(sol);
This enables us to perform operations on
the solution y1 of the initial value
problem. For instance, we can evaluate y1
at a given value of t, or we can plot y1 on
some given t interval.
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> evalf(subs(t=4,y1));
Example 1: A 1st order linear equation
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plot(y1,t=0..20);
Example 2: A 2nd order homogeneous
linear equation
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Solve the equation y’’ - 2y’ - 4y = 0.
Find the general solution and also the solution
that satisfies the initial conditions y(0) = 2, y’(0)
= -7/4.
SOLUTION: Again the first step is to enter the
differential equation. In the following command
note that the first term the second derivative is
entered using t$2. This causes the differentiation
operation to be executed twice.
eq2 := diff(y(t),t$2) - 2*diff(y(t),t) - 4*y(t) = 0;
dsolve(eq2,y(t));
Example 2: A 2nd order linear equation
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This is the general solution. Note that it contains
two exponential terms, each multiplied by an
arbitrary constant. One of the exponentials is
positive, and so grows with increasing t, while
the other is negative, and decays as t increases.
Now we solve the initial value problem. To enter
the second initial condition, denote y’(0) by
D(y)(0). We will assign the name y2 to this
solution, combining the dsolve and rhs
commands.
> y2 := rhs(dsolve({eq2,y(0) = 2,D(y)(0)=7/4},y(t)));
Example 3: A 2nd order nonhomogeneous linear equation
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Find the general solution of the equation y’’ + 4
y’ + 5 y = 3 exp(-t) + 4 sin(3t).
Also find the solution satisfying the initial
conditions
y(0) = 3/2, y’(0) = 2, and plot its graph.
SOLUTION. As in the preceding examples we find
the general solution by entering the equation and
then using dsolve.
> eq3
:=diff(y(t),t$2)+4*diff(y(t),t)+5*y(t)=3*exp(t)+4*sin(3*t);
Example 3: A 2nd order nonhomogeneous linear equation
sol := dsolve(eq3,y(t));
 init := y(0) = 3/2, D(y)(0) = 2;
 sol := dsolve({eq3,init},y(t)):
 plot(y3,t=0..15);
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Example 4: Find the general solution of
a system
Find the general solution of the system
dx/dt = -2 x + y, dy/dt = x - 2 y.
 Also find the solution that satisfies the
initial conditions x(0) = 0, y(0) = 2.
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sys := diff(x(t),t) = -2*x(t)+y(t), diff(y(t),t) =
x(t) - 2*y(t);
dsolve({sys},{x(t),y(t)});
Example 4: Find the general solution of
a system
Here is the solution of the initial value
problem.
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dsolve({sys,x(0)=0,y(0)=2},{x(t),y(t)});
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Ass 2
In each of the following problems use
dsolve to find the general solution of the
given differential equation. If initial
conditions are given, also find the solution
that satisfies them.
 Plot this solution.
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Ass 2 cont.
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1. dy/dt = t + exp(-2t) + 3y, y(0) = 2
2. dy/dt = 3 sin(t) + 1 - y, y(0) = -1
3. y’’ + (5/2)y’ + y = 0, y(0) = 2, y’(0) = 3
4. y’’ + 6y’ + 10y = 0, y(0) = 1, y’(0) = 1
5. y’’ + (5/2)y’ + y = 3 cos(2t) + 6 exp(-2t), y(0) = 2, y’(0)
=3
6. y’’ + 6y’ + 10y= 5 exp(-3t)cos(t) + 2 exp(-t)cos(3t),
y(0) = 1, y’(0) = 2
7. y’’’ + 3y’’ + 3y’ + 2y = 0,y(0) = 2, y’(0) = -1, y’’(0) = 3
8. y’’’’ + 4y’’’ + 10y’’ + 11y’ + 10y = 0, y(0) = 2, y’(0) = 3,
y’’(0) = -1, y’’’(0) = 0
9. y’’’ + 3y’’ + 3y’ + 2y = exp(-t)sin(t) + 3t^2 - 4t + 5, y(0)
= 2, y’(0) = -1, y’’(0) = 3
10. dx/dt = 3x - 2y, dy/dt = 2x - 2y,x(0) = 3, y(0) = 2