Difference between revisions of "Maple/Differential Equations"
(→Making a Plot) |
|||
Line 67: | Line 67: | ||
=== Solve Equation / Initial Conditions === | === Solve Equation / Initial Conditions === | ||
+ | Without initial conditions, you can use | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | soln1 := dsolve([eqn1], [y(t)]) | ||
+ | </syntaxhighlight> | ||
+ | to get:<center><math>\mathit{soln1}:=\left\{y\! \left(t\right)=-\frac{1}{2} g \,t^{2}+\textit{_}\mathit{C1} t+\textit{_}\mathit{C2}\right\}</math></center> | ||
+ | |||
+ | You can also put in symbolic initial conditions. To put in a derivative condition, use the format <code>D^n(var)(time)=val</code> to establish a condition for the ''n''th derivative of <code>var</code> at time <code>time</code>. For instance, if you know some initial velocity and position, we can use: | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | soln2 := dsolve([eqn1, y(0) = y0, D(y)(0) = v0], [y(t)]) | ||
+ | </syntaxhighlight>to get:<center><math>\mathit{soln2}:=y\! \left(t\right)=-\frac{1}{2} g \,t^{2}+\mathit{v0} t+\mathit{y0}</math></center> | ||
+ | |||
+ | |||
+ | === Making a Plot === | ||
+ | Since your answer has symbols, you will need to replace them with numbers before plotting: | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | plot(subs(soln2, g = 9.81, y0 = 5, v0 = 10, y(t)), t = 0 .. 2) | ||
+ | </syntaxhighlight> | ||
+ | and can of course add other plotting options as needed. Once again, note that ''soln2'' is a single equation, not a collection, so there is no need to de-bracket. | ||
+ | |||
+ | |||
+ | == More Complicated Coupled Example == | ||
+ | For a series RLC circuit with an applied voltage $$v_S(t)$$ across the series network, you can find a coupled set of differential equations using KVL and the model equation for the capacitor, respectively, as:<center><math> | ||
+ | \begin{align*} | ||
+ | L\frac{iL(t)}{st}+R\,i_L(t)+v_C(t)&=0\\ | ||
+ | i_L(t)&=C\frac{dv_C(t)}{dt} | ||
+ | \end{align*}</math></center> | ||
+ | We can '''try''' to solve these symbolically, but the results will not be particularly helpful. | ||
+ | |||
+ | === Initialization === | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | restart | ||
+ | </syntaxhighlight> | ||
+ | |||
+ | === Define Equations === | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | eqn1 := -vs(t) + L*diff(iL(t), t) + R*iL(t) + vC(t) = 0; | ||
+ | eqn2 := iL(t) = C*diff(vC(t), t); | ||
+ | eqns := [eqn1, eqn2] | ||
+ | </syntaxhighlight> | ||
+ | |||
+ | === Solve Equation / Initial Conditions === | ||
+ | === No Great Version 1 ==== | ||
+ | If you try | ||
+ | <syntaxhighlight lang=Maple> | ||
+ | soln1 := dsolve(eqns, [iL(t), vC(t)]) | ||
+ | </syntaxhighlight> | ||
+ | you will get: | ||
+ | <center><math> | ||
+ | \mathit{iL}\! \left(t\right) = \left(\frac{{\mathrm e}^{-\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}} C R}{2 \sqrt{C \left(R^{2} C-4 L\right)}\, L}+\frac{{\mathrm e}^{-\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}}{2 L}\right) \left(gray}{\int}\mathit{vs}\! \left(t\right) {\mathrm e}^{\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}{d}t\right)+\left(-\frac{{\mathrm e}^{-\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}} C R}{2 \sqrt{C \left(R^{2} C-4 L\right)}\, L}+\frac{{\mathrm e}^{-\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}}{2 L}\right) \left({\int}\mathit{vs}\! \left(t\right) {\mathrm e}^{\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}{d}t\right)+\left(-\frac{C^{2} R^{2}}{2 \sqrt{C \left(R^{2} C-4 L\right)}\, L}-\frac{R C}{2 L}+\frac{2 C}{\sqrt{C \left(R^{2} C-4 L\right)}}\right) \textit{_}\mathit{C1} {\mathrm e}^{-\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}+\left(\frac{C^{2} R^{2}}{2 \sqrt{C \left(R^{2} C-4 L\right)}\, L}-\frac{R C}{2 L}-\frac{2 C}{\sqrt{C \left(R^{2} C-4 L\right)}}\right) \textit{_}\mathit{C2} {\mathrm e}^{-\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}, | ||
+ | \\ | ||
+ | \mathit{vC}\! \left(t\right)=-\frac{-{\mathrm e}^{-\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}} \textit{_}\mathit{C2} \sqrt{C \left(R^{2} C-4 L\right)}-{\mathrm e}^{-\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}} \textit{_}\mathit{C1} \sqrt{C \left(R^{2} C-4 L\right)}+\left({\int}\mathit{vs}\! \left(t\right) {\mathrm e}^{\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}{d}t\right) {\mathrm e}^{-\frac{R t}{L}+\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}-\left({\int}\mathit{vs}\! \left(t\right) {\mathrm e}^{\frac{\left(R C-\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}{d}t\right) {\mathrm e}^{-\frac{R t}{L}+\frac{\left(R C+\sqrt{C \left(R^{2} C-4 L\right)}\right) t}{2 L C}}}{\sqrt{C \left(R^{2} C-4 L\right)}}\mathrm{ | ||
+ | \\} | ||
+ | </math></center> | ||
+ | |||
Without initial conditions, you can use | Without initial conditions, you can use | ||
<syntaxhighlight lang=Maple> | <syntaxhighlight lang=Maple> |
Revision as of 22:51, 26 February 2024
Contents
Introduction
This page focuses on using Maple to find both the symbolic and the numeric solutions to differential equations obtained from electric circuits.
Note: There is a...decidedly more complicated explanation of these things at Maple/Differential Equations/Old; the goal for Spring 2024 and beyond is to keep things simpler.
Very Basic Example
Imagine you have the equation $$2\frac{dy(t)}{dt} + 3 y(t) = 4$$ with the initial condition $$y(0)=5$$, and you want to solve for $$y(t)$$. You can do this with Maple as follows:
Initialization
As always, it is a good idea to include the restart
command in your worksheet:
restart
Define Equation
In the same way that you were able to assign a linear algebra expression to a variable, you can do the same with a differential equation. The key is to note that the Maple diff
command can be used to calculate or represent a derivative. You will need to explicitly let Maple know that your variable is a function (in our case, a function of $$t$$) by including that parameter with the variable. Given that, you can store the differential equation in a variable called eqn
with:
eqn1:=2*diff(y(t), t)+3*y(t)=4
Solve Equation
Solving a system of differential equations is also similar to solving a system of linear algebra equations - the main differences are that you will use dsolve
instead of solve
, you must continue to use $$y(t)$$ instead of just $$y$$, and you may end up needing to add some initial conditions. The code
soln1:=dsolve([eqn1], [y(t)])
will result in a solution of:
Initial Condition
To incorporate initial conditions, you will give the dsolve
command information about the value of the variable (or, for higher order differential equations, the value and values of the derivatives of the variable). For example, to solve our sample equation with $$y(0)=5$$, you will include the initial condition by adding y(0)=5
to the equations:
soln2:=dsolve([eqn1, y(0)=5], [y(t)])
will produce
Note that the initial condition does not have to be at time 0; if you know that $$y(6)=7$$, you can use that as well:
soln3:=dsolve([eqn1, y(6)=7], [y(t)])
will produce
Making a Plot
Once you have a solution or set of solutions, you can plot them using subs. For instance, to plot $$y(t)$$ in soln2 above (where we set $$y(0)=5$$) you can use:
plot(subs(soln2, y(t)), t = 0 .. 5)
and can of course add other plotting options as needed. Note that soln2 is a single equation, not a collection. If you end up getting a collection of results, you may need to de-bracket them.
Second Order Example
Now imagine that you want to solve for $$y(t)$$ in
We can solve this using all symbols or numbers, and we can solve with or without initial conditions.
Initialization
restart
Define Equation
eqn1 := diff(y(t), t, t) = -g
Solve Equation / Initial Conditions
Without initial conditions, you can use
soln1 := dsolve([eqn1], [y(t)])
to get:
You can also put in symbolic initial conditions. To put in a derivative condition, use the format D^n(var)(time)=val
to establish a condition for the nth derivative of var
at time time
. For instance, if you know some initial velocity and position, we can use:
soln2 := dsolve([eqn1, y(0) = y0, D(y)(0) = v0], [y(t)])
to get:
Making a Plot
Since your answer has symbols, you will need to replace them with numbers before plotting:
plot(subs(soln2, g = 9.81, y0 = 5, v0 = 10, y(t)), t = 0 .. 2)
and can of course add other plotting options as needed. Once again, note that soln2 is a single equation, not a collection, so there is no need to de-bracket.
More Complicated Coupled Example
For a series RLC circuit with an applied voltage $$v_S(t)$$ across the series network, you can find a coupled set of differential equations using KVL and the model equation for the capacitor, respectively, as:
We can try to solve these symbolically, but the results will not be particularly helpful.
Initialization
restart
Define Equations
eqn1 := -vs(t) + L*diff(iL(t), t) + R*iL(t) + vC(t) = 0;
eqn2 := iL(t) = C*diff(vC(t), t);
eqns := [eqn1, eqn2]
Solve Equation / Initial Conditions
No Great Version 1 =
If you try
soln1 := dsolve(eqns, [iL(t), vC(t)])
you will get:
Without initial conditions, you can use
soln1 := dsolve([eqn1], [y(t)])
to get:
You can also put in symbolic initial conditions. To put in a derivative condition, use the format D^n(var)(time)=val
to establish a condition for the nth derivative of var
at time time
. For instance, if you know some initial velocity and position, we can use:
soln2 := dsolve([eqn1, y(0) = y0, D(y)(0) = v0], [y(t)])
to get:
Making a Plot
Since your answer has symbols, you will need to replace them with numbers before plotting:
plot(subs(soln2, g = 9.81, y0 = 5, v0 = 10, y(t)), t = 0 .. 2)
and can of course add other plotting options as needed. Once again, note that soln2 is a single equation, not a collection, so there is no need to de-bracket.