Error using sym>convertChar (line 1537) when solving an ODE using laplace transform. How to fix the problem?

2021 4 30
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更新时间：2021 5 17
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%%CODE
%% Solving 2nd order ODE using laplace transform
clc, clear , close all
syms x t s X F
F = laplace('diff(x(t),t,t)+7*diff(x(t),t)+10*x(t)= 20',s); % solving using laplace transform
F = subs(F,{'x(0)','D(x)(0)'},{5,3}); % initial values
F = subs(F,{'laplace(x(t),t,s)'},{X}); % substituting the initial values then solve Laplace
X = solve(F,'X');
X = ilaplace(X);
X = simplify(X); pretty(X);
disp(X)
%% ERROR BELOW
Error using sym>convertChar (line 1537)
Character vectors and strings in the first argument can only
specify a variable or number. To evaluate character vectors and
strings representing symbolic expressions, use 'str2sym'.
Error in sym>tomupad (line 1253)
S = convertChar(x);
Error in sym (line 220)
S.s = tomupad(x);
Error in transform (line 22)
if ~isa(f, 'sym'), f = sym(f); end
Error in sym/laplace (line 28)
L = transform('symobj::laplace', 't', 's', 'z', F, varargin{:});
Error in HW_1_3_2_4070H300 (line 4)
F = laplace('diff(x(t),t,t)+7*diff(x(t),t)+10*x(t)= 20',s); %
solving using laplace transform

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Walter Roberson
Walter Roberson 2021-5-15
Since R2017b (I think it is) you cannot pass character vectors to laplace() . You need to construct the symbolic equation and pass that instead.

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 采纳的回答

Star Strider
Star Strider 2021-5-15
Ran in:

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Ran in:
Eliminate the single quotes, use double equal signs in the symbolic expression, express ‘x’ as ‘x(t)’ in the syms declaration (otherwise, ‘x’ is assumed to be a constant), and it works —
syms x(t) t s X F
Dx = diff(x);
D2x = diff(Dx);
F = laplace(D2x+7*Dx+10*x(t) == 20,s); % solving using laplace transform
F = subs(F,{x(0),Dx(0)},{5,3}); % initial values
F = subs(F,{laplace(x(t),t,s)},{X}); % substituting the initial values then solve Laplace
X = solve(F,X);
X = ilaplace(X);
X = simplify(X); pretty(X);
exp(-2 t) 6 - exp(-5 t) 3 + 2
disp(X)
Character arrays have not been allowed for the last few releases. (The one exception that remains is in the sym funciton.)

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Konard Adams
Konard Adams 2021-5-17
编辑：Walter Roberson 2021-5-17
Firstly thank you
I have notived t is not needed ..Why so?
%% Solving 2nd order ODE using laplace transform
clc, clear , close all
% syms x(t) t s X F
syms x(t) s X F
Dx = diff(x);
D2x = diff(Dx);
F = laplace(D2x+7*Dx+10*x(t) == 20,s); % solving using laplace transform
F = subs(F,{x(0),Dx(0)},{5,3}); % initial values
F = subs(F,{laplace(x(t),s)},{X}); % substituting the initial values then solve Laplace
%F = subs(F,{laplace(x(t),t,s)},{X}); % t is not really needed
X = solve(F,X);
disp('laplace solution = '),disp(X)
X = ilaplace(X);
X = simplify(X); %pretty(X);
disp('inverse laplace Solution = '),disp(X)
Star Strider
Star Strider 2021-5-17
As always, my pleasure!
Declaring ‘t’ specifically is not absolutely required (in this instance), since it is implied in declaring ‘x(t)’. It is needed in that declaration because ‘x(t)’ needs to be defined as a function in the syms declaration in order for the code using it to work (specifically with respect to defining and calculating the detrivatives of ‘x(t)’).

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Walter Roberson
Walter Roberson 2021-5-15
编辑：Walter Roberson 2021-5-15
Ran in:

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Ran in:
%%CODE
%% Solving 2nd order ODE using laplace transform
syms x(t) s X F
Dx = diff(x,t);
D2x = diff(Dx,t);
eqn = D2x + 7*Dx + 10*x(t) == 20
eqn(t) = 
F = laplace(eqn,s) % solving using laplace transform
F = 
F = subs(F,{x(0), Dx(0)},{5,3}) % initial values
F = 
F = subs(F, {laplace(x(t),t,s)},{X}) % substituting the initial values then solve Laplace
F = 
X = solve(F, X)
X = 
X = ilaplace(X)
X = 

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Konard Adams
Konard Adams 2021-5-17
Firstly thank you
I have notived t is not needed ..Why so?
%% Solving 2nd order ODE using laplace transform
clc, clear , close all
% syms x(t) t s X F
syms x(t) s X F
Dx = diff(x);
D2x = diff(Dx);
F = laplace(D2x+7*Dx+10*x(t) == 20,s); % solving using laplace transform
F = subs(F,{x(0),Dx(0)},{5,3}); % initial values
F = subs(F,{laplace(x(t),s)},{X}); % substituting the initial values then solve Laplace
%F = subs(F,{laplace(x(t),t,s)},{X}); % t is not really needed
X = solve(F,X);
disp('laplace solution = '),disp(X)
X = ilaplace(X);
X = simplify(X); %pretty(X);
disp('inverse laplace Solution = '),disp(X)

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