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本集的数值案例如下：
Example 1:
syms p x r
solve(p*sin(x) == r)
%chooses 'x' as the unknown and returns
ans =
asin(r/p)
pi - asin(r/p)
Example 2:
syms x y
[Sx,Sy] = solve(x^2 + x*y + y == 3,x^2 - 4*x + 3 == 0)
%returns
Sx =
1
3
Sy =
1
-3/2
Example 3:
syms x y
S = solve(x^2*y^2 - 2*x - 1 == 0,x^2 - y^2 - 1 == 0)
%returns
%the solutions in a structure.
S =
x: [8x1 sym]
y: [8x1 sym]
Example 4:
syms a u v
[Su,Sv] = solve(a*u^2 + v^2 == 0,u - v == 1)
%regards 'a' as a
%parameter and solves the two equations for u and v.
Example 5:
syms a u v w
S = solve(a*u^2 + v^2,u - v == 1,a,u)
%regards 'v' as a
%parameter, solves the two equations, and returns S.a and S.u.
%When assigning the result to several outputs, the order in which
%the result is returned depends on the order in which the variables
%are given in the call to solve:
[U,V] = solve(u + v,u - v == 1, u, v)
%assigns the value for u to U
%and the value for v to V. In contrast to that
[U,V] = solve(u + v,u - v == 1, v, u)
%assigns the value for v to U
%and the value of u to V.
Example 6:
syms a u v
[Sa,Su,Sv] = solve(a*u^2 + v^2,u - v == 1,a^2 - 5*a + 6)
%solves
%the three equations for a, u and v.
Example 7:
syms x
S = solve(x^(5/2) == 8^(sym(10/3)))
%returns all three complex solutions:
S =
16
- 4*5^(1/2) - 4 + 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i
- 4*5^(1/2) - 4 - 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i
Example 8:
syms x
S = solve(x^(5/2) == 8^(sym(10/3)), 'PrincipalValue', true)
%selects one of these:
S =
- 4*5^(1/2) - 4 + 4*2^(1/2)*(5 - 5^(1/2))^(1/2)*i
Example 9:
syms x
S = solve(x^(5/2) == 8^(sym(10/3)), 'IgnoreAnalyticConstraints', true)
%ignores branch cuts during internal simplifications and, in this case,
%also returns only one solution:
S =
16
Example 10:
syms x
S = solve(sin(x) == 0) returns 0
S = solve(sin(x) == 0, 'ReturnConditions', true)
%returns a structure expressing
%the full solution:
S.x = k*pi
S.parameters = k
S.conditions = in(k, 'integer')
Example 11:
syms x y real
[S, params, conditions] = solve(x^(1/2) = y, x, 'ReturnConditions', true)
%assigns solution, parameters and conditions to the outputs.
%In this example, no new parameters are needed to express the solution:
S =
y^2
params =
Empty sym: 1-by-0
conditions =
0 <= y
Example 12:
syms a x y
[x0, y0, params, conditions] = solve(x^2+y, x, y, 'ReturnConditions', true)
%generates a new parameter z to express the infinitely many solutions.
%This z can be any complex number, both solutions are valid without
%restricting conditions:
x0 =
-(-z)^(1/2)
(-z)^(1/2)
y0 =
z
z
params =
z
conditions =
true
true
Example 13:
syms t positive
solve(t^2-1)
ans =
1
solve(t^2-1, 'IgnoreProperties', true)
ans =
1
-1
Example 14:
solve(x^3-1) returns all three complex roots:
ans =
1
- 1/2 + (3^(1/2)*i)/2
- 1/2 - (3^(1/2)*i)/2
solve(x^3-1, 'Real', true) only returns the real root:
ans =
1例1:解超越方程组.
e1 = sym('x^x-4=0');
e2 = sym('2*x*y+x=1');
[x,y] = solve(e1,e2);
fprintf('x = %f \n',x);
fprintf('y = %f \n',y);
%%结果
%x = 2.000000
%y = -0.250000 例2：解非线性方程组.
e1 = sym('x+y+z=u');
e2 = sym('2*x*z-y*u=-1');
e3 = sym('(x+y)^2-z=u');
e4 = sym('x*u+y*z=4');
[x,y,z,u] = solve(e1,e2,e3,e4);
disp(double(x));
disp(double(y));
disp(double(z));
disp(double(u));