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

2 次查看（过去 30 天）
Konard Adams
Konard Adams 2021-4-30
编辑： Walter Roberson 2021-5-17
%%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

请先登录，再回答此问题。

采纳的回答

Star Strider
Star Strider 2021-5-15
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.)
  4 个评论
显示 2更早的评论

请先登录，再进行评论。

更多回答（1 个）

Walter Roberson
Walter Roberson 2021-5-15
编辑：Walter Roberson 2021-5-15
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 = 
  2 个评论
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)

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Equation Solving 的更多信息

标签

产品


版本

R2020a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by