Solving Equations
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Solving Algebraic Equations
If S is a symbolic expression,
solve(S)
attempts to find values of the symbolic variable in S (as determined by symvar) for which S is zero. For example,
syms a b c x S = a*x^2 + b*x + c; solve(S)
uses the familiar quadratic formula to produce
ans = -(b + (b^2 - 4*a*c)^(1/2))/(2*a) -(b - (b^2 - 4*a*c)^(1/2))/(2*a)
This is a symbolic vector whose elements are the two solutions.
If you want to solve for a specific variable, you must specify that variable as an additional argument. For example, if you want to solve S for b, use the command
b = solve(S,b)
which returns
b = -(a*x^2 + c)/x
Note that these examples assume equations of the form f(x) = 0. If you need to solve equations of the form f(x) = q(x), you must use quoted strings. In particular, the command
s = solve('cos(2*x) + sin(x) = 1')returns a vector with three solutions.
s =
0
pi/6
(5*pi)/6There are also solutions at each of these results plus kπ for integer k, as you can see in the MuPAD solution:
Several Algebraic Equations
This section explains how to solve systems of equations using Symbolic Math Toolbox software. As an example, suppose you have the system
and you want to solve for x and y. First, create the necessary symbolic objects.
syms x y;
alpha = sym('alpha');There are several ways to address the output of solve. One is to use a two-output call
[x, y] = solve(x^2*y^2, x-y/2 - alpha)
which returns
x =
alpha
0
y =
0
-2*alphaModify the first equation to x2y2 = 1 and there are more solutions.
eqs1 = 'x^2*y^2=1, x-y/2-alpha'; [x,y] = solve(eqs1)
produces four distinct solutions:
x = alpha/2 + (alpha^2 + 2)^(1/2)/2 alpha/2 + (alpha^2 - 2)^(1/2)/2 alpha/2 - (alpha^2 + 2)^(1/2)/2 alpha/2 - (alpha^2 - 2)^(1/2)/2 y = (alpha^2 + 2)^(1/2) - alpha (alpha^2 - 2)^(1/2) - alpha - alpha - (alpha^2 + 2)^(1/2) - alpha - (alpha^2 - 2)^(1/2)
Since you did not specify the dependent variables, solve uses symvar to determine the variables.
This way of assigning output from solve is quite successful for "small" systems. Plainly, if you had, say, a 10-by-10 system of equations, typing
[x1,x2,x3,x4,x5,x6,x7,x8,x9,x10] = solve(...)
is both awkward and time consuming. To circumvent this difficulty, solve can return a structure whose fields are the solutions. In particular, consider the system u^2 - v^2 = a^2, u + v = 1, a^2 - 2*a = 3. The command
S = solve('u^2 - v^2 = a^2', 'u + v = 1', 'a^2 - 2*a = 3')returns
S =
a: [2x1 sym]
u: [2x1 sym]
v: [2x1 sym]The solutions for a reside in the "a-field" of S. That is,
S.a
produces
ans = -1 3
Similar comments apply to the solutions for u and v. The structure S can now be manipulated by field and index to access a particular portion of the solution. For example, if you want to examine the second solution, you can use the following statement
s2 = [S.a(2), S.u(2), S.v(2)]
to extract the second component of each field.
s2 = [ 3, 5, -4]
The following statement
M = [S.a, S.u, S.v]
creates the solution matrix M
M = [ -1, 1, 0] [ 3, 5, -4]
whose rows comprise the distinct solutions of the system.
Linear systems of simultaneous equations can also be solved using matrix division. For example,
clear u v x y syms u v x y S = solve(x + 2*y - u, 4*x + 5*y - v); sol = [S.x; S.y] A = [1 2; 4 5]; b = [u; v]; z = A\b
results in
sol =
(2*v)/3 - (5*u)/3
(4*u)/3 - v/3
z =
(2*v)/3 - (5*u)/3
(4*u)/3 - v/3Thus s and z produce the same solution, although the results are assigned to different variables.
Single Differential Equation
The function dsolve computes symbolic solutions to ordinary differential equations. The equations are specified by symbolic expressions containing the letter D to denote differentiation. The symbols D2, D3, ... DN, correspond to the second, third, ..., Nth derivative, respectively. Thus, D2y is the toolbox equivalent of d2y/dt2. The dependent variables are those preceded by D and the default independent variable is t. Note that names of symbolic variables should not contain D. The independent variable can be changed from t to some other symbolic variable by including that variable as the last input argument.
Initial conditions can be specified by additional equations. If initial conditions are not specified, the solutions contain constants of integration, C1, C2, etc.
The output from dsolve parallels the output from solve. That is, you can call dsolve with the number of output variables equal to the number of dependent variables or place the output in a structure whose fields contain the solutions of the differential equations.
Example 1
The following call to dsolve
dsolve('Dy = t*y')uses y as the dependent variable and t as the default independent variable.
The output of this command is
ans = C2*exp(t^2/2)
y = C*exp(t^2/2) is a solution to the equation for any constant C.
To specify an initial condition, use
y = dsolve('Dy = t*y', 'y(0) = 2')This produces
y = 2*exp(t^2/2)
Notice that y is in the MATLAB workspace, but the independent variable t is not. Thus, the command diff(y,t) returns an error. To place t in the workspace, enter syms t.
Example 2
Nonlinear equations may have multiple solutions, even when initial conditions are given:
x = dsolve('(Dx + x)^2 = 1', 'x(0) = 0')results in
x = 1/exp(t) - 1 1 - 1/exp(t)
Example 3
Here is a second-order differential equation with two initial conditions, and the default independent variable changed to x. The commands
y = dsolve('D2y = cos(2*x) - y', 'y(0) = 1', 'Dy(0) = 0', 'x');
simplify(y)produce
ans = 1 - (8*(cos(x)/2 - 1/2)^2)/3
Example 4
The key issues in this example are the order of the equation and the initial conditions. To solve the ordinary differential equation
with x as the independent variable, type
u = dsolve('D3u = u',...
'u(0) = 1', 'Du(0) = -1', 'D2u(0) = pi', 'x')Use D3u to represent d3u/dx3 and D2u(0) for 
.
.u = (pi*exp(x))/3 - (cos((3^(1/2)*x)/2)*(pi/3 - 1))/exp(x/2) ... - (3^(1/2)*sin((3^(1/2)*x)/2)*(pi + 1))/(3*exp(x/2))
Further ODE Examples
This table shows a few more examples of differential equations and their Symbolic Math Toolbox syntax. The final entry in the table is the Airy differential equation, whose solution is referred to as the Airy function.
Differential Equation | MATLAB Command |
|---|---|
 y(0) = 1 | |
2x2y′′ + 3xy′ – y = 0 ( ′ = d/dx) | y = dsolve('2*x^2*D2y + 3*x*Dy - y = 0', 'x') |
  (The Airy equation) |
Several Differential Equations
The function dsolve can handle several ordinary differential equations in several variables, with or without initial conditions. For example, here is a pair of linear, first-order equations.
S = dsolve('Df = 3*f + 4*g', 'Dg = -4*f + 3*g')The toolbox returns the computed solutions in the structure S. You can determine the values of f and g by typing
f = S.f g = S.g
f = C2*cos(4*t)*exp(3*t) + C1*sin(4*t)*exp(3*t) g = C1*cos(4*t)*exp(3*t) - C2*sin(4*t)*exp(3*t)
If you prefer to recover f and g directly, as well as include initial conditions, type
[f, g] = dsolve('Df = 3*f + 4*g, Dg = -4*f + 3*g',...
'f(0) = 0, g(0) = 1')f = sin(4*t)*exp(3*t) g = cos(4*t)*exp(3*t)
Now, suppose you are solving a system of differential equations in matrix form. For example, solve the system Y′ = AY + B, where A, B, and Y represent the following matrices:
syms t x y; A = [1 2; -1 1]; B = [1; t]; Y = [x; y]; sys = A*Y + B
sys = x + 2*y + 1 t - x + y
The dsolve function does not accept matrices. To be able to use this solver, extract the components of the matrix and convert them to strings:
eq1 = char(sys(1)) eq2 = char(sys(2))
eq1 = x + 2*y + 1 eq2 = t - x + y
Use the strcat function to concatenate the left and right sides of the equations. Use the dsolve function to solve the system:
[x, y] = dsolve(strcat('Dx = ',eq1), strcat('Dy = ',eq2))x = 2^(1/2)*exp(t)*cos(2^(1/2)*t)*(C6 + (4*sin(2^(1/2)*t) +... 2^(1/2)*cos(2^(1/2)*t) + 6*t*sin(2^(1/2)*t) +... 6*2^(1/2)*t*cos(2^(1/2)*t))/(18*exp(t))) +... 2^(1/2)*exp(t)*sin(2^(1/2)*t)*(C5 - (4*cos(2^(1/2)*t) -... 2^(1/2)*sin(2^(1/2)*t) + 6*t*cos(2^(1/2)*t) -... 6*2^(1/2)*t*sin(2^(1/2)*t))/(18*exp(t))) y = exp(t)*cos(2^(1/2)*t)*(C5 - (4*cos(2^(1/2)*t) -... 2^(1/2)*sin(2^(1/2)*t) + 6*t*cos(2^(1/2)*t) -... 6*2^(1/2)*t*sin(2^(1/2)*t))/(18*exp(t))) -... exp(t)*sin(2^(1/2)*t)*(C6 + (4*sin(2^(1/2)*t) +... 2^(1/2)*cos(2^(1/2)*t) + 6*t*sin(2^(1/2)*t) +... 6*2^(1/2)*t*cos(2^(1/2)*t))/(18*exp(t)))
