Equations and systems solver - MATLAB solve - MathWorks France (2024)

Solve the quadratic equation without specifying a variable to solve for. solve chooses x to return the solution.

syms a b c xeqn = a*x^2 + b*x + c == 0

eqn =$$a\hspace{0.17em}{x}^2+b\hspace{0.17em}x+c=0$$

S = solve(eqn)

S =
$$\left(\begin{array}{c}-\frac{b+\sqrt{b^2-4\hspace{0.17em}a\hspace{0.17em}c}}{2\hspace{0.17em}a}\\ -\frac{b-\sqrt{b^2-4\hspace{0.17em}a\hspace{0.17em}c}}{2\hspace{0.17em}a}\end{array}\right)$$

Specify the variable to solve for and solve the quadratic equation for a.

Sa = solve(eqn,a)

Sa =
$$-\frac{c+b\hspace{0.17em}x}{x^2}$$

Solve a fifth-degree polynomial. It has five solutions.

syms xeqn = x^5 == 3125;S = solve(eqn,x)

S =
$$\begin{array}{l}\left(\begin{array}{c}5\\ -{\sigma }_1-\frac{5}{4}-\frac{5\hspace{0.17em}\sqrt{2}\hspace{0.17em}\sqrt{5-\sqrt{5}}\hspace{0.17em}\mathrm{i}}{4}\\ -{\sigma }_1-\frac{5}{4}+\frac{5\hspace{0.17em}\sqrt{2}\hspace{0.17em}\sqrt{5-\sqrt{5}}\hspace{0.17em}\mathrm{i}}{4}\\ {\sigma }_1-\frac{5}{4}-\frac{5\hspace{0.17em}\sqrt{2}\hspace{0.17em}\sqrt{\sqrt{5}+5}\hspace{0.17em}\mathrm{i}}{4}\\ {\sigma }_1-\frac{5}{4}+\frac{5\hspace{0.17em}\sqrt{2}\hspace{0.17em}\sqrt{\sqrt{5}+5}\hspace{0.17em}\mathrm{i}}{4}\end{array}\right)\\ \\ \mathrm{where}\\ \\ \mathrm{}{\sigma }_1=\frac{5\hspace{0.17em}\sqrt{5}}{4}\end{array}$$

Return only real solutions by setting 'Real' option to true. The only real solutions of this equation is 5.

S = solve(eqn,x,'Real',true)

S =$$5$$

When solve cannot symbolically solve an equation, it tries to find a numeric solution using vpasolve. The vpasolve function returns the first solution found.

Try solving the following equation. solve returns a numeric solution because it cannot find a symbolic solution.

syms xeqn = sin(x) == x^2 - 1;S = solve(eqn,x)

Warning: Unable to solve symbolically. Returning a numeric solution using <a href="matlab:web(fullfile(docroot, 'symbolic/vpasolve.html'))">vpasolve</a>.

S =$$-0.63673265080528201088799090383828$$

Plot the left and the right sides of the equation. Observe that the equation also has a positive solution.

fplot([lhs(eqn) rhs(eqn)], [-2 2])

Find the other solution by directly calling the numeric solver vpasolve and specifying the interval.

V = vpasolve(eqn,x,[0 2])

V =$$1.4096240040025962492355939705895$$

When solving for multiple variables, it can be more convenient to store the outputs in a structure array than in separate variables. The solve function returns a structure when you specify a single output argument and multiple outputs exist.

Solve a system of equations to return the solutions in a structure array.

syms u veqns = [2*u + v == 0, u - v == 1];S = solve(eqns,[u v])

S = struct with fields: u: 1/3 v: -2/3

Access the solutions by addressing the elements of the structure.

S.u

ans =
$$\frac{1}{3}$$

S.v

ans =
$$-\frac{2}{3}$$

Using a structure array allows you to conveniently substitute solutions into other expressions.

Use the subs function to substitute the solutions S into other expressions.

expr1 = u^2;e1 = subs(expr1,S)

e1 =
$$\frac{1}{9}$$

expr2 = 3*v + u;e2 = subs(expr2,S)

e2 =
$$-\frac{5}{3}$$

If solve returns an empty object, then no solutions exist.

eqns = [3*u+2, 3*u+1];S = solve(eqns,u)

 S = Empty sym: 0-by-1 

The solve function can solve inequalities and return solutions that satisfy the inequalities. Solve the following inequalities.

$$x>0$$

$$y>0$$

$$x^2+y^2+xy<1$$

Set 'ReturnConditions' to true to return any parameters in the solution and conditions on the solution.

syms x yeqn1 = x > 0;eqn2 = y > 0;eqn3 = x^2 + y^2 + x*y < 1;eqns = [eqn1 eqn2 eqn3];S = solve(eqns,[x y],'ReturnConditions',true);S.x

ans =
$$\frac{\sqrt{u-3\hspace{0.17em}{v}^2}}{2}-\frac{v}{2}$$

S.y

ans =$$v$$

S.parameters

ans =$$\left(\begin{array}{cc}u& v\end{array}\right)$$

S.conditions

ans =$$4\hspace{0.17em}{v}^2<u\wedge u<4\wedge 0<v$$

The parameters u and v do not exist in MATLAB® workspace and must be accessed using S.parameters.

Check if the values u = 7/2 and v = 1/2 satisfy the condition using subs and isAlways.

condWithValues = subs(S.conditions, S.parameters, [7/2,1/2]);isAlways(condWithValues)

ans = logical 1

isAlways returns logical 1 (true) indicating that these values satisfy the condition. Substitute these parameter values into S.x and S.y to find a solution for x and y.

xSol = subs(S.x, S.parameters, [7/2,1/2])

xSol =
$$\frac{\sqrt{11}}{4}-\frac{1}{4}$$

ySol = subs(S.y, S.parameters, [7/2,1/2])

ySol =
$$\frac{1}{2}$$

Solve the system of equations.

$$2u^2+v^2=0$$

$$u-v=1$$

When solving for more than one variable, the order in which you specify the variables defines the order in which the solver returns the solutions. Assign the solutions to variables solv and solu by specifying the variables explicitly. The solver returns an array of solutions for each variable.

syms u veqns = [2*u^2 + v^2 == 0, u - v == 1];vars = [v u];[solv, solu] = solve(eqns,vars)

solv =
$$\left(\begin{array}{c}-\frac{2}{3}-\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\\ -\frac{2}{3}+\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\end{array}\right)$$

solu =
$$\left(\begin{array}{c}\frac{1}{3}-\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\\ \frac{1}{3}+\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\end{array}\right)$$

Entries with the same index form the pair of solutions.

solutions = [solv solu]

solutions =
$$\left(\begin{array}{cc}-\frac{2}{3}-\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}& \frac{1}{3}-\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\\ -\frac{2}{3}+\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}& \frac{1}{3}+\frac{\sqrt{2}\hspace{0.17em}\mathrm{i}}{3}\end{array}\right)$$

Return the complete solution of an equation with parameters and conditions of the solution by specifying 'ReturnConditions' as true.

Solve the equation $$\sin (x)=0$$. Provide two additional output variables for output arguments parameters and conditions.

syms xeqn = sin(x) == 0;[solx,parameters,conditions] = solve(eqn,x,'ReturnConditions',true)

solx =$$\pi \hspace{0.17em}k$$

parameters =$$k$$

conditions =$$k\in \mathbb{Z}$$

The solution $$\pi k$$ contains the parameter $$k$$, where $$k$$ must be an integer. The variable $$k$$ does not exist in the MATLAB® workspace and must be accessed using parameters.

Restrict the solution to $$0<x<2\pi$$. Find a valid value of $$k$$ for this restriction. Assume the condition, conditions, and use solve to find $$k$$. Substitute the value of $$k$$ found into the solution for $$x$$.

assume(conditions)restriction = [solx > 0, solx < 2*pi];solk = solve(restriction,parameters)

solk =$$1$$

valx = subs(solx,parameters,solk)

valx =$$\pi$$

Alternatively, determine the solution for $$x$$ by choosing a value of $$k$$. Check if the value chosen satisfies the condition on $$k$$ using isAlways.

Check if $$k=4$$ satisfies the condition on $$k$$.

condk4 = subs(conditions,parameters,4);isAlways(condk4)

ans = logical 1

isAlways returns logical 1(true), meaning that 4 is a valid value for $$k$$. Substitute $$k$$ with 4 to obtain a solution for $$x$$. Use vpa to obtain a numeric approximation.

valx = subs(solx,parameters,4)

valx =$$4\hspace{0.17em}\pi$$

vpa(valx)

ans =$$12.5664$$

Solve the equation $\exp \left(\log \left(\mathit{x}\right)\log \left(3\mathit{x}\right)\right)=4$.

By default, solve does not apply simplifications that are not valid for all values of $$x$$. In this case, the solver does not assume that $$x$$ is a positive real number, so it does not apply the logarithmic identity $$\log (3x)=\log (3)+\log (x)$$. As a result, solve cannot solve the equation symbolically.

syms xeqn = exp(log(x)*log(3*x)) == 4;S = solve(eqn,x)

Warning: Unable to solve symbolically. Returning a numeric solution using <a href="matlab:web(fullfile(docroot, 'symbolic/vpasolve.html'))">vpasolve</a>.

S =$$-14.009379055223370038369334703094-2.9255310052111119036668717988769\hspace{0.17em}\mathrm{i}$$

Set 'IgnoreAnalyticConstraints' to true to apply simplification rules that might allow solve to find a solution. For details, see Algorithms.

S = solve(eqn,x,'IgnoreAnalyticConstraints',true)

S =
$$\left(\begin{array}{c}\frac{\sqrt{3}\hspace{0.17em}{\mathrm{e}}^{-\frac{\sqrt{\log \left(256\right)+{\log \left(3\right)}^2}}{2}}}{3}\\ \frac{\sqrt{3}\hspace{0.17em}{\mathrm{e}}^{\frac{\sqrt{\log \left(256\right)+{\log \left(3\right)}^2}}{2}}}{3}\end{array}\right)$$

solve applies simplifications that allow the solver to find a solution. The mathematical rules applied when performing simplifications are not always valid in general. In this example, the solver applies logarithmic identities with the assumption that $$x$$ is a positive real number. Therefore, the solutions found in this mode should be verified.

The sym and syms functions let you set assumptions for symbolic variables.

Assume that the variable x is positive.

syms x positive

When you solve an equation for a variable under assumptions, the solver only returns solutions consistent with the assumptions. Solve this equation for x.

eqn = x^2 + 5*x - 6 == 0;S = solve(eqn,x)

S =$$1$$

Allow solutions that do not satisfy the assumptions by setting 'IgnoreProperties' to true.

S = solve(eqn,x,'IgnoreProperties',true)

S =
$$\left(\begin{array}{c}-6\\ 1\end{array}\right)$$

For further computations, clear the assumption that you set on the variable x by recreating it using syms.

syms x

When you solve a polynomial equation, the solver might use root to return the solutions. Solve a third-degree polynomial.

syms x aeqn = x^3 + x^2 + a == 0;solve(eqn, x)

ans =
$$\left(\begin{array}{c}\mathrm{root}\left(z^3+z^2+a,z,1\right)\\ \mathrm{root}\left(z^3+z^2+a,z,2\right)\\ \mathrm{root}\left(z^3+z^2+a,z,3\right)\end{array}\right)$$

Try to get an explicit solution for such equations by calling the solver with 'MaxDegree'. The option specifies the maximum degree of polynomials for which the solver tries to return explicit solutions. The default value is 2. Increasing this value, you can get explicit solutions for higher order polynomials.

Solve the same equations for explicit solutions by increasing the value of 'MaxDegree' to 3.

S = solve(eqn, x, 'MaxDegree', 3)

S =
$$\begin{array}{l}\left(\begin{array}{c}\frac{1}{9\hspace{0.17em}{\sigma }_1}+{\sigma }_1-\frac{1}{3}\\ -\frac{1}{18\hspace{0.17em}{\sigma }_1}-\frac{{\sigma }_1}{2}-\frac{1}{3}-\frac{\sqrt{3}\hspace{0.17em}\left(\frac{1}{9\hspace{0.17em}{\sigma }_1}-{\sigma }_1\right)\hspace{0.17em}\mathrm{i}}{2}\\ -\frac{1}{18\hspace{0.17em}{\sigma }_1}-\frac{{\sigma }_1}{2}-\frac{1}{3}+\frac{\sqrt{3}\hspace{0.17em}\left(\frac{1}{9\hspace{0.17em}{\sigma }_1}-{\sigma }_1\right)\hspace{0.17em}\mathrm{i}}{2}\end{array}\right)\\ \\ \mathrm{where}\\ \\ \mathrm{}{\sigma }_1={\left(\sqrt{{\left(\frac{a}{2}+\frac{1}{27}\right)}^2-\frac{1}{729}}-\frac{a}{2}-\frac{1}{27}\right)}^{1/3}\end{array}$$

Solve the equation $$\sin (x)+\cos (2x)=1$$.

Instead of returning an infinite set of periodic solutions, the solver picks three solutions that it considers to be the most practical.

syms xeqn = sin(x) + cos(2*x) == 1;S = solve(eqn,x)

S =
$$\left(\begin{array}{c}0\\ \frac{\pi }{6}\\ \frac{5\hspace{0.17em}\pi }{6}\end{array}\right)$$

Choose only one solution by setting 'PrincipalValue' to true.

S1 = solve(eqn,x,'PrincipalValue',true)

S1 =$$0$$

S — Solutions of equation
symbolic array

Solutions of an equation, returned as a symbolic array. The size of a symbolic array corresponds to the number of the solutions.

Y — Solutions of system of equations
structure

Solutions of a system of equations, returned as a structure. The number of fields in the structure correspond to the number of independent variables in a system. If 'ReturnConditions' is set to true, the solve function returns two additional fields that contain the parameters in the solution, and the conditions under which the solution is true.

y1,...,yN — Solutions of system of equations
symbolic variables

Solutions of a system of equations, returned as symbolic variables. The number of output variables or symbolic arrays must be equal to the number of independent variables in a system. If you explicitly specify independent variables vars, then the solver uses the same order to return the solutions. If you do not specify vars, the toolbox sorts independent variables alphabetically, and then assigns the solutions for these variables to the output variables.

parameters — Parameters in solution
vector of generated parameters

conditions — Conditions under which solutions are valid
vector of symbolic expressions