Equations and systems solver

String inputs will be removed in a future release. Instead,
use `syms`

to declare variables
and replace inputs such as `solve('2*x == 1','x')`

with ```
solve(2*x
== 1,x)
```

.

uses
additional options specified by one or more `S`

= solve(`eqn`

,`var`

,`Name,Value`

)`Name,Value`

pair
arguments.

solves
the system of equations `Y`

= solve(`eqns`

,`vars`

)`eqns`

for the variables `vars`

and
returns a structure that contains the solutions. If you do not specify `vars`

, `solve`

uses `symvar`

to find the variables to solve
for. In this case, the number of variables that `symvar`

finds
is equal to the number of equations `eqns`

.

uses
additional options specified by one or more `Y`

= solve(`eqns`

,`vars`

,`Name,Value`

)`Name,Value`

pair
arguments.

`[`

solves
the system of equations `y1,...,yN`

] = solve(`eqns`

,`vars`

)`eqns`

for the variables `vars`

.
The solutions are assigned to the variables `y1,...,yN`

.
If you do not specify the variables, `solve`

uses `symvar`

to
find the variables to solve for. In this case, the number of variables
that `symvar`

finds is equal to the number of output
arguments `N`

.

`[`

uses
additional options specified by one or more `y1,...,yN`

] = solve(`eqns`

,`vars`

,`Name,Value`

)`Name,Value`

pair
arguments.

`[`

returns
the additional arguments `y1,...,yN`

,`parameters`

,`conditions`

]
= solve(`eqns`

,`vars`

,'`ReturnConditions`

',true)`parameters`

and `conditions`

that
specify the parameters in the solution and the conditions on the solution.

Use the `==`

operator to specify
the equation `sin(x) == 1`

and solve it.

syms x eqn = sin(x) == 1; solx = solve(eqn,x)

solx = pi/2

Find the complete solution of the same equation by specifying
the `ReturnConditions`

option as `true`

.
Specify output variables for the solution, the parameters in the solution,
and the conditions on the solution.

[solx, params, conds] = solve(eqn, x, 'ReturnConditions', true)

solx = pi/2 + 2*pi*k params = k conds = in(k, 'integer')

The solution `pi/2 + 2*pi*k`

contains the parameter `k`

which
is valid under the condition `in(k, 'integer')`

.
This condition means the parameter `k`

must be an
integer.

If `solve`

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

returns an empty
object with a warning, solutions might exist but `solve`

did
not find any solutions.

solve(3*x+2, 3*x+1, x)

ans = Empty sym: 0-by-1

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 x [solx, param, cond] = solve(sin(x) == 0, x, 'ReturnConditions', true)

solx = pi*k param = k cond = in(k, 'integer')

The solution `pi*k`

contains the parameter `k`

and
is valid under the condition `in(k,'integer')`

. This
condition means the parameter `k`

must be an integer. `k`

does
not exist in the MATLAB^{®} workspace and must be accessed using `param`

.

Find a valid value of `k`

for ```
0 <
x < 2*pi
```

by assuming the condition, `cond`

,
and using `solve`

to solve these conditions for `k`

.
Substitute the value of `k`

found into the solution
for `x`

.

assume(cond) solk = solve([solx > 0, solx < 2*pi], param) valx = subs(solx, param, solk)

solk = 1 valx = pi

A valid value of `k`

for ```
0 < x
< 2*pi
```

is `1`

. This produces the value ```
x
= pi
```

.

Alternatively, find a 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`

.

isAlways(subs(cond, param, 4))

ans = 1

`isAlways`

returns logical `1`

(`true`

),
meaning `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, param, 4) vpa(valx)

valx = 4*pi ans = 12.566370614359172953850573533118

Avoid ambiguities when solving equations with
symbolic parameters by specifying the variable for which you want
to solve an equation. If you do not specify the variable, `solve`

chooses
a variable using `symvar`

. First, solve the quadratic
equation without specifying a variable. `solve`

chooses *x* to
return the familiar solution. Then solve the quadratic equation for `a`

to
return the solution for `a`

.

syms a b c x sol = solve(a*x^2 + b*x + c == 0) sola = solve(a*x^2 + b*x + c == 0, a)

sol = -(b + (b^2 - 4*a*c)^(1/2))/(2*a) -(b - (b^2 - 4*a*c)^(1/2))/(2*a) sola = -(c + b*x)/x^2

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.

Solve this system of equations and 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 v [solv, solu] = solve([2*u^2 + v^2 == 0, u - v == 1], [v, u])

solv = - (2^(1/2)*1i)/3 - 2/3 (2^(1/2)*1i)/3 - 2/3 solu = 1/3 - (2^(1/2)*1i)/3 (2^(1/2)*1i)/3 + 1/3

Entries with the same index form the solutions of a system.

solutions = [solv, solu]

solutions = [ - (2^(1/2)*1i)/3 - 2/3, 1/3 - (2^(1/2)*1i)/3] [ (2^(1/2)*1i)/3 - 2/3, (2^(1/2)*1i)/3 + 1/3]

A solution of the system is ```
v = - (2^(1/2)*1i)/3 -
2/3
```

, and `u = 1/3 - (2^(1/2)*1i)/3`

.

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

S = u: [1x1 sym] v: [1x1 sym]

Access the solutions by addressing the elements of the structure.

S.u S.v

ans = 1/3 ans = -2/3

Using a structure array allows you to conveniently substitute
solutions into expressions. The `subs`

function
substitutes the correct values irrespective of which variables you
substitute.

Substitute solutions into expressions using the structure `S`

.

subs(u^2, S) subs(3*v+u, S)

ans = 1/9 ans = -5/3

Return the complete solution of a system of
equations with parameters and conditions of the solution by specifying `ReturnConditions`

as `true`

.

syms x y S = solve([sin(x)^2 == cos(y), 2*x == y],... [x, y], 'ReturnConditions', true); S.x S.y S.conditions S.parameters

ans = pi*k - asin(3^(1/2)/3) asin(3^(1/2)/3) + pi*k ans = 2*pi*k - 2*asin(3^(1/2)/3) 2*asin(3^(1/2)/3) + 2*pi*k ans = in(k, 'integer') in(k, 'integer') ans = k

A solution is formed by the elements of the same index in `S.x`

, `S.y`

,
and `S.conditions`

. Any element of `S.parameters`

can
appear in any solution. For example, a solution is ```
x = pi*k
- asin(3^(1/2)/3)
```

, and `y = 2*pi*k - 2*asin(3^(1/2)/3)`

,
with the parameter `k`

under the condition ```
in(k,
'integer')
```

. This condition means `k`

must
be an integer for the solution to be valid. `k`

does
not exist in the MATLAB workspace and must be accessed with `S.parameters`

.

For the first solution, find a valid value of `k`

for ```
0
< x < pi
```

by assuming the condition `S.conditions(1)`

and
using `solve`

to solve these conditions for `k`

.
Substitute the value of `k`

found into the solution
for `x`

.

assume(S.conditions(1)) solk = solve([S.x(1) > 0, S.x(1) < pi], S.parameters) solx = subs(S.x(1), S.parameters, solk)

solk = 1 solx = pi - asin(3^(1/2)/3)

A valid value of `k`

for ```
0 < x
< pi
```

is `1`

. This produces the value ```
x
= pi - asin(3^(1/2)/3)
```

.

Alternatively, find a 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`

.

isAlways(subs(S.conditions(1), S.parameters, 4))

ans = 1

`isAlways`

returns logical `1`

(`true`

)
meaning `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(S.x(1), S.parameters, 4) vpa(valx)

valx = 4*pi - asin(3^(1/2)/3) ans = 11.950890905688785612783108943994

Try solving the following equation. The symbolic solver cannot find an exact symbolic solution for this equation, and therefore issues a warning before calling the numeric solver. Because the equation is not polynomial, an attempt to find all possible solutions can take a long time. The numeric solver does not try to find all numeric solutions for this equation. Instead, it returns only the first solution it finds.

syms x solve(sin(x) == x^2 - 1, x)

Warning: Cannot solve symbolically. Returning a numeric approximation instead. > In solve at 301 ans = -0.63673265080528201088799090383828

Plot the left and the right sides of the equation in one graph. The graph shows that the equation also has a positive solution.

ezplot(sin(x), -2, 2) hold on ezplot(x^2 - 1, -2, 2) hold off

Find this solution by calling the numeric solver `vpasolve`

directly
and specifying the interval where this solution can be found.

vpasolve(sin(x) == x^2 - 1, x, [0 2])

ans = 1.4096240040025962492355939705895

`solve`

can solve inequalities
to find a solution that satisfies the inequalities.

Solve the following inequalities. Set `ReturnConditions`

to `true`

to
return any parameters in the solution and conditions on the solution.

$$\begin{array}{l}x>0\\ y>0\\ {x}^{2}+{y}^{2}+xy<1\end{array}$$

syms x y S = solve(x^2 + y^2 + x*y < 1, x > 0, y > 0,... [x, y], 'ReturnConditions', true); solx = S.x soly = S.y params = S.parameters conditions = S.conditions

solx = (- 3*v^2 + u)^(1/2)/2 - v/2 soly = v params = [ u, v] conditions = 4*v^2 < u & u < 4 & 0 < v

The parameters `u`

and `v`

do
not exist in the 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`

.

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

ans = 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`

.

solx = subs(S.x, S.parameters, [7/2,1/2]) soly = subs(S.y, S.parameters, [7/2,1/2])

solx = 11^(1/2)/4 - 1/4 soly = 1/2

Convert the solution into numeric form by using `vpa`

.

vpa(solx) vpa(soly)

ans = 0.57915619758884996227873318416767 ans = 0.5

Solve this equation. It has five solutions.

syms x solve(x^5 == 3125, x)

ans = 5 - (2^(1/2)*(5 - 5^(1/2))^(1/2)*5i)/4 - (5*5^(1/2))/4 - 5/4 (2^(1/2)*(5 - 5^(1/2))^(1/2)*5i)/4 - (5*5^(1/2))/4 - 5/4 (5*5^(1/2))/4 - (2^(1/2)*(5^(1/2) + 5)^(1/2)*5i)/4 - 5/4 (5*5^(1/2))/4 + (2^(1/2)*(5^(1/2) + 5)^(1/2)*5i)/4 - 5/4

Return only real solutions by setting argument `Real`

to `true`

.
The only real solution of this equation is `5`

.

solve(x^5 == 3125, x, 'Real', true)

ans = 5

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

syms x solve(sin(x) + cos(2*x) == 1, x)

ans = 0 pi/6 (5*pi)/6

Pick only one solution using `PrincipalValue`

.

solve(sin(x) + cos(2*x) == 1, x, 'PrincipalValue', true)

ans = 0

Try to solve this equation. By default, `solve`

does
not apply simplifications that are not always mathematically correct.
As a result, `solve`

cannot solve this equation
symbolically.

syms x solve(exp(log(x)*log(3*x)) == 4, x)

Warning: Cannot solve symbolically. Returning a numeric approximation instead. ans = - 14.009379055223370038369334703094 - 2.9255310052111119036668717988769i

Set `IgnoreAnalyticConstraints`

to `true`

to
apply simplifications that might allow `solve`

to
find a result. For details, see Algorithms.

S = solve(exp(log(x)*log(3*x)) == 4, x, 'IgnoreAnalyticConstraints', true)

S = (3^(1/2)*exp(-(log(256) + log(3)^2)^(1/2)/2))/3 (3^(1/2)*exp((log(256) + log(3)^2)^(1/2)/2))/3

`solve`

applies simplifications that allow
it to find a solution. The simplifications applied do not always hold.
Thus, the solutions in this mode might not be correct or complete,
and need verification.

The `sym`

and `syms`

functions
let you set assumptions for symbolic variables.

Assume that the variable *x* can have only
positive values.

syms x positive

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

.

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

ans = 1

Allow solutions that do not satisfy the assumptions by setting `IgnoreProperties`

to `true`

.

solve(x^2 + 5*x - 6 == 0, x, 'IgnoreProperties', true)

ans = -6 1

For further computations, clear the assumption that you set
on the variable *x*.

syms x clear

When solving polynomials, `solve`

might
return solutions containing `RootOf`

. To numerically
approximate these solutions, use `vpa`

. Consider
the following equation and solution.

syms x s = solve(x^4 + x^3 + 1 == 0, x)

s = root(z^4 + z^3 + 1, z, 1) root(z^4 + z^3 + 1, z, 2) root(z^4 + z^3 + 1, z, 3) root(z^4 + z^3 + 1, z, 4)

Because there are no parameters in this solution, use `vpa`

to
approximate it numerically.

vpa(s)

ans = - 1.0189127943851558447865795886366 + 0.60256541999859902604398442197193i - 1.0189127943851558447865795886366 - 0.60256541999859902604398442197193i 0.5189127943851558447865795886366 + 0.666609844932018579153758800733i 0.5189127943851558447865795886366 - 0.666609844932018579153758800733i

When you solve a higher order polynomial equation,
the solver might use `RootOf`

to return the results.
Solve an equation of order 4.

syms x a solve(x^4 + x^3 + a == 0, x)

ans = root(z^4 + z^3 + a, z, 1) root(z^4 + z^3 + a, z, 2) root(z^4 + z^3 + a, z, 3) root(z^4 + z^3 + a, z, 4)

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 a third order polynomial by increasing the value of `MaxDegree`

to `3`

to
return explicit solutions instead of `RootOf`

.

S = solve(x^3 + x + a == 0, x, 'MaxDegree', 3); pretty(S)

/ 1 \ | #1 - ---- | | 3 #1 | | | | / 1 \ | | sqrt(3) | ---- + #1 | 1i | | 1 \ 3 #1 / #1 | | ---- - ------------------------ - -- | | 6 #1 2 2 | | | | / 1 \ | | sqrt(3) | ---- + #1 | 1i | | \ 3 #1 / 1 #1 | | ------------------------ + ---- - -- | \ 2 6 #1 2 / where / / 2 \ \1/3 | | a 1 | a | #1 == | sqrt| -- + -- | - - | \ \ 4 27 / 2 /

Was this topic helpful?