”=” is an assignment operator; it fixes the LHS to be equal to the RHS. A variable can be assigned many types of values: Numbers, functions, lists, even whole equations.
z = 1
p = a x^2 + b x + c
rep = {a -> 1, b -> 2, c -> 3}
”==” is a comparison operator; it compares the LHS to the RHS and returns “true” or “false”. == can be used in conjunction with = to assign an equation to a single variable.
z == 1
z == 2
equation = x^2 == y
Solve[equation, x] finds the solution to an equation, or a system of equations contained in a list:
Solve[a x^2 + b x + c == 0, x]
Solve[{x^2 + x == 2 w, 3 x^2 - 2 == w}, {x, w}]
(Note that the output of Solve is a list of replacement rules, not an assignment.)
The number of equations must be equal to the number of variables. To reduce the equations by removing one variable, use Eliminate[{eq1, eq2}, x]:
Eliminate[{x^2 + x == 2 w, 3 x^2 - 2 == w}, x]
For differential equations, DSolve[eq1, x[t], t] finds the general solution x[t]
(You can use an apostrophe in a differential equation instead of the extraneous D[x[t], t]
DSolve[x''[t] + \[Omega]^2 x[t] == 0, x[t], t]
It also works for a system of equations, just like Solve:
DSolve[{x''[t] + \[Omega]1^2 x[t] == 0,
y'[t] + \[Omega]2^2 x[t] == 0}, {x[t], y[t]}, t]
The general solution contains arbitrary constants C[n]. For boundary value problems, include boundary conditions!
DSolve[{x''[t] + \[Omega]^2 x[t] == 0, x[0] == 1, x'[1] == 2}, x[t], t]
If equations don’t have an exact solution (or if it takes forever to find one), use NSolve[func[x], x]. The argument must be fully numerical (have no arbitrary parameters), except for the variable x!
NSolve[LegendreP[20, 2, x] == 0, x]
Similarly, with differential equations, but in this case you need to specify boundary conditions:
sol = NDSolve[{6 y'[x] - 2 y[x] - x y[x]^4 == 0, y[0] == -2},
y[x], {x, -10, 10}]
Plot[y[x] /.sol, {x, -10, 10}]