How linear system of equations can be solved in matlab

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Wajahat
Wajahat 2018-9-8
编辑： madhan ravi 2018-9-10
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".
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Wajahat
Wajahat 2018-9-10
@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
Walter Roberson 2018-9-10
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.

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madhan ravi
madhan ravi 2018-9-10
编辑：madhan ravi 2018-9-10
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

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