A matlab solution to equation 50

Solving Equations with MATLAB: Two Steps, One Note.

Step 1: Use "syms" to define variables;

Step 2: Use the "solve" function to solve the equation;

Note: When expressing the function, use "==, such as x+y==5 (instead of x+y=5);

Example: syms x y z ; solve(Equation 1, Equation 2, Equation 3, x,y,z) <>

In MATLAB, we can solve this equation by defining a family equation and using a numerical solver. Knowing the first and second derivatives, we can express the equation as an ordinary differential equation (ODE). To solve for x(t), we can use MATLAB's ODE45 function.

First, the equation is represented in MATLAB:

function dxdt = myode(t, x, f0, omega_l, omega)

dxdt = zeros(2,1);

dxdt(1) =x(2);

dxdt(2) =f0 * cos(omega * t) -f0/omega * sin(omega_l * t) -sin(omega_l * ti));

endNext, define the known parameters:

f0 = 1;% instead of a specific value.

omega_l = 1;% instead of a specific value.

omega = 1;% instead of a specific value.

ti = 1;% instead of a specific value.

Now, use the ode45 function to solve for ode:

tspan = 0, 10];Instead of a specific time frame.

x0 = 0; 0];Initial conditions (position and speed grip).

t, x] =ode45(@(t, x) myode(t, x, f0, omega_l, omega), tspan, x0);

Here, tspan is the timeframe and x0 is the initial condition. The ODE45 function will return the solution of the time vector t and position (x(1)) and velocity (x(2)).

Note: The deficit variable is used in the expression, as this is the standard notation for MATLAB. You can modify the variable name to meet your requirements.

How to use MATLAB programming to find the solution of the equation in the graph, for the equation in the graph to solve the solve() and vpasolve() functions, it does not give us all the solutions. So how do you get it?

First, let's try to plot the graph using the plot function (ezplot, plot), and we can see that the equation has two solutions, i.e., x has a solution near 0 and .

Then, use the vpasolve function to find them one by one.

Implementation: syms x

ezplot(sin(x)-x 2 2,[-5,5]).

grid on %.

x1 = vpasolve(sin(x)-x 2 2 == 0, x, 0) % to find the solution of x near 0.

x2 = vpasolve(sin(x)-x 2 2 == 0, x, find the solution where x is nearby.

How to solve the following implicit trigonometric equations with MATLAB. Generally, it can be solved using the vpasolve function. For example, 6 = 2 and 7 = 6, then the following ** is used.

theta6=pi/2;theta7=pi/6;

syms a1 % declares variables.

alpha1=vpasolve(6/sin(a1)==7/(pi-(theta6-theta7+a1)),a1])

alpha=double(alpha1*180 pi) % radian conversion angle.

If 6 and 7 are a series of values, you can use a vector array to find their corresponding alpha values with a loop statement. Rule.

theta6=[。theta7=[。

for i=1:n %n is the number of 6 and 7.

syms a1 % declares variables.

alpha1=vpasolve(6/sin(a1)==7/(pi-(theta6(i)-theta7(i)+a1)),a1])

alpha(i)=double(alpha1*180/pi)

end<>

The landlord should use less matlab:

The first equation is missing a closing parenthesis;

When using the solve solver, the first thing to do is to define a symbolic variable with the syms command, which in this case is preceded by it.

syms pp qq;

The solve command can only recognize numeric or symbolic variables, and cannot recognize constants such as oc and os defined by the landlord, if the landlord also defines them as symbolic variables, the result will be represented by these variables, for example: pp = oc + os;

The equal sign "=" is the assignment command, and "==" should be used to express the equation when using the solve command.

The solve command is not a panacea, and when the equation is very complex (as in this case), you often don't get an exact answer, and after correcting all the mistakes in this example, you still can't get a satisfactory exact solution (of course, it may be that I am not good at it...). Instead, the result is two empty symbolic variables: pp=[empty sym] qq=

Provide a simple solution example for the landlord for reference:

clc;clear

syms x y a b

x, y] = solve(x + y == a, x - y == b, x, y)

The result is:

x = a/2 + b/2

y = a/2 - b/2