How linear system of equations can be solved in matlab

How we can solve following linear system of equations in matlab?

A1_{x}=1i*a*(A1+A2);

A2_{x}=1i*a*(A1-A2);

A1_{t}=(-1i*a./2)*A1-A2;

A2_{t}=A1+(1i*a./2)*A2;

where A1=A1(x,t) and A2=A2(x,t) and "a" is an arbitrary constant. How can these equations can be solved in matlab?

And A1_{x} means partial derivative of A1 w.r.t "x".

[madhan ravi]

Can you provide all the details and in latex form ?

[Wajahat]

Latex form available @ https://math.stackexchange.com/questions/2909818/how-we-can-solve-linear-system-of-equations

[madhan ravi]

Do you have symbolic and partial differential equations toolbox?

[Wajahat]

@Ravi, for these differential equations, i have no symbolic tool box

[madhan ravi]

if you want to solve it symbolically then you should have those toolboxes otherwise you can use ode solvers to solve it numerically

[Wajahat]

Can you solve it symbolically, I am new in matlab and can't solve it symbolically...

[Wajahat]

@Ravi, I have try to solve it symbolically, but matlab shows an error.

syms l p q

syms f1(x) f2(x)

S = dsolve(diff(f1) == l.^{-1}.*1i.*p.*f1 + l.^{-1}.*1i.*q.*f2, diff(f2) == l.^{-1}.*1i.*q.*f1 - l.^{-1}.*1i.*p.*f2);

S.f1

S.f2

Can you remove an error

[Walter Roberson]

What is the I.^{-1} intended to mean?

It is not possible to raise anything to a cell array, not unless you define your own object class and override the power() operator.

Answer 1

[madhan ravi]

Try this @Wajahat:

syms l p q

syms f1(x) f2(x)

%edited after sir Walters comment 
S1 = diff(f1) == l.^(-1).*1i.*p.*f1 + l.^(-1).*1i.*q.*f2;
S2 = diff(f2) == l.^{-1}.*1i.*q.*f1 - l.^{-1}.*1i.*p.*f2;

S = dsolve(S1,S2)

S.f2

S.f1

[madhan ravi]

@Wajahat if it’s working please accept the answer