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

%%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

[Naveed Naqvi]

Having the same exact problem unfortunately :/

[Walter Roberson]

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

Answer 1

[Star Strider]

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.)

[Konard Adams]

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]

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)’).

[Konard Adams]

Understood, cheers!

[Star Strider]

Cheers!

Answer 2

[Walter Roberson]

%%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 = 

[Konard Adams]

Cheers! I now understand , great explanation.

[Konard Adams]

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)