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

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Konard Adams on 30 Apr 2021
Edited: Walter Roberson on 17 May 2021
%%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'.
S = convertChar(x);
Error in sym (line 220)
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
##### 2 CommentsShowHide 1 older comment
Walter Roberson on 15 May 2021
Since R2017b (I think it is) you cannot pass character vectors to laplace() . You need to construct the symbolic equation and pass that instead.

Star Strider on 15 May 2021
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.)
Star Strider on 17 May 2021
Cheers!

Walter Roberson on 15 May 2021
Edited: Walter Roberson on 15 May 2021
%%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 =
##### 2 CommentsShowHide 1 older comment
Konard Adams on 17 May 2021
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)

R2020a

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