solving second order differential equations with LaPlace Transforms: getting an error when trying to input an initial condition (R2017b) ("error: untitled line")

This is my code. I am getting an error in

 syms s 
eqn=sym('D(D(y))(t)+D(y)(t)+y(t)=((t+1)^3)*exp(-t)*cos(t)*sin(3*t)');
lteqn=laplace(eqn,t,s)
syms y ;
neweqn=subs(lteqn,{'laplace(y(t),t,s)','y(0)',''D(y)(0)'},{Y,1,0})
% the line above has the error, and the issue is with the 'D(y)(0)' term
ytrans=simplify(solve(neweqn,Y))
y=ilaplace(ytrans,s,t)

Answer 1

[Star Strider]

The laplace function does not accept initial conditions (as dsolve does).

I requested this as an enhancement several months ago. It has thus far not appeared.

You will have to do coding gymnastics with the subs function to do the necessary substitutions.

Answer 2

[Sulaymon Eshkabilov]

%% Here is the 3rd solution that works in MATLAB2018a/b and 2019a
clc; clearvars; syms  t s Y y(t) Dy(t)
assume([t Y] > 0);
Dy = diff(y, t);
D2y = diff(Dy, t);
LS = ((t+1)^3)*exp(-t)*cos(t)*sin(3*t);
% Diff. Equation formulation:
EQN=D2y+Dy+y-LS;
% Laplace transform of the Diff. Equation
LEQN=laplace(EQN,t,s);  
% Substitute ICs and initiate the arbitrary unknown "Y"
LT_Y=subs(LEQN,laplace(y,t,s),Y);
LT_Y=subs(LT_Y, y(0), 1);                            %  y(0) = 1
LT_Y=subs(LT_Y, subs(diff(y(t), t), t, 0), 0);  %  dy(0)= 0
% Solve for the arbitrary unknown: Y
ys=solve(LT_Y,Y);
% Inverse of the Laplace Transform:
y=ilaplace(ys,s,t)
ezplot(y); shg

[Yusif Razzaq]

really helped me out. thanks

Answer 3

[Sulaymon Eshkabilov]

% Here are two solutions:
%% Solution 1. Note: laplace() takes ICs.
clearvars; syms s t y Y
eqn=sym('D(D(y))(t)+D(y)(t)+y(t)=((t+1)^3)*exp(-t)*cos(t)*sin(3*t)'); % Works Matlab 2017a/b or earlier vers not for 2018a or later vers.
lteqn=laplace(eqn,t,s);
neweqn=subs(lteqn,{'laplace(y(t),t,s)','y(0)','D(y)(0)'},{Y,1,0})
% No error now
ytrans=simplify(solve(neweqn,Y))
y=ilaplace(ytrans,s,t), ezplot(y), shg

%% Solution 2
clearvars; syms t y(t)
%assume([t Y] > 0)
Dy = diff(y);
D2y = diff(y, 2);
EQN = D2y+Dy+y-((t+1)^3)*exp(-t)*cos(t)*sin(3*t);
y_s = dsolve(EQN, y(0)==1, Dy(0)==0);
ezplot(y_s), legend('toggle'), hold off; shg

[Thilini Madhushika]

I also have same problem. I don't know how to substitute initial conditions for dy1 and dy2.

k=2,m=1 for both equations. condition 1 =1,condition2=1,condition3=sqrt(6),condition4= -1*sqrt(6)

This is my code:

clc
clear all
syms y1(t) y2(t) s
assume(t>0)
dy1 = diff(y1,t);
dy2 = diff(y2,t);
disp('assume(t>0)')
disp('diff(y1,t)=dy1')
disp('diff(y2,t)=dy2')
disp('diff(y1,t,2) == (-2*(k/m)*y1+(k/m)*y2)')
disp('Enter the coefficients correctly for the first equation')
k=input('k=');
m=input('m=');
eqn1= diff(y1,t,2) == (-2*(k/m)*y1+(k/m)*y2);
disp('diff(y2,t,2) == ((k/m)*y1-2*(k/m)*y2)')
k=input('k=');
m=input('m=');
disp('Enter the coefficients correctly for the second equation')
eqn2= diff(y2,t,2) == ((k/m)*y1-2*(k/m)*y2);
c1=input('enter condition 1:');
c2=input('enter condition 2:');
c3=input('enter condition 3:');
c4=input('enter condition 4:');
eqn1LT = laplace(eqn1,t,s)
eqn2LT = laplace(eqn2,t,s)
syms y1_LT y2_LT
eqn1LT = subs(eqn1LT,[laplace(y1,t,s) laplace(y2,t,s)],[y1_LT y2_LT])
eqn2LT = subs(eqn2LT,[laplace(y1,t,s) laplace(y2,t,s)],[y1_LT y2_LT])
eqns = [eqn1LT eqn2LT]
vars = [y1_LT y2_LT]
[y1_LT,y2_LT] = solve(eqns,vars)
y1sol = ilaplace(y1_LT,s,t)
y2sol = ilaplace(y2_LT,s,t)
y1sol = simplify(y1sol)
y2sol = simplify(y2sol)
vars = [y1(0) y2(0) dy1(0) dy2(0)]
values = [c1 c2 c3 c4]
y1sol = subs(y1sol,vars,values)
y2sol = subs(y2sol,vars,values)
%disp('Enter the range for the plot the graph')
%tmin = input('tmin value:');
%tmax = input('tmax value:');
subplot(2,1,1)
ezplot(y1sol,t)
title('The Graph Of y1(t) Vs t')
ylabel('y1(t)')
xlabel('t')
subplot(2,1,2)
ezplot(y2sol,t)
title('The Graph Of y2(t) Vs t')
ylabel('y2(t)')
xlabel('t')

[Walter Roberson]

@Thilini Madhushika

You look like you are substituting for dy1(0) and dy2(0) already ? Not sure what you want to do differently ?

if ~isunix()
    k1 = input('first k=');
    m1 = input('first m=');
    
    k2 = input('second k=');
    m2 = input('second m=');

    c1 = input('enter condition 1:');
    c2 = input('enter condition 2:');
    c3 = input('enter condition 3:');
    c4 = input('enter condition 4:');
else
    k1 = 2;
    m1 = 1;
    k2 = 2;
    m2 = 1;
    
    c1 = 1;
    c2 = 1;
    c3 = sqrt(6);
    c4 = -1*sqrt(6);
end

syms y1(t) y2(t) s
assume(t>0)
dy1 = diff(y1,t);
dy2 = diff(y2,t);
disp('assume(t>0)')

assume(t>0)

disp('diff(y1,t)=dy1')

diff(y1,t)=dy1

disp('diff(y2,t)=dy2')

diff(y2,t)=dy2

disp('diff(y1,t,2) == (-2*(k/m)*y1+(k/m)*y2)')

diff(y1,t,2) == (-2*(k/m)*y1+(k/m)*y2)

disp('Enter the coefficients correctly for the first equation')

Enter the coefficients correctly for the first equation

k = k1; m = m1;

eqn1= diff(y1,t,2) == (-2*(k/m)*y1+(k/m)*y2);
disp('diff(y2,t,2) == ((k/m)*y1-2*(k/m)*y2)')

diff(y2,t,2) == ((k/m)*y1-2*(k/m)*y2)

k = k2; m = m2;

disp('Enter the coefficients correctly for the second equation')

Enter the coefficients correctly for the second equation

eqn2= diff(y2,t,2) == ((k/m)*y1-2*(k/m)*y2);

eqn1LT = laplace(eqn1,t,s)

eqn1LT = 

eqn2LT = laplace(eqn2,t,s)

eqn2LT = 

syms y1_LT y2_LT
eqn1LT = subs(eqn1LT,[laplace(y1,t,s) laplace(y2,t,s)],[y1_LT y2_LT])

eqn1LT = 

eqn2LT = subs(eqn2LT,[laplace(y1,t,s) laplace(y2,t,s)],[y1_LT y2_LT])

eqn2LT = 

eqns = [eqn1LT eqn2LT]

eqns = 

vars = [y1_LT y2_LT]

vars = 

[y1_LT,y2_LT] = solve(eqns,vars)

y1_LT = 

y2_LT = 

y1sol = ilaplace(y1_LT,s,t)

y1sol = 

y2sol = ilaplace(y2_LT,s,t)

y2sol = 

y1sol = simplify(y1sol)

y1sol = 

y2sol = simplify(y2sol)

y2sol = 

vars = [y1(0) y2(0) dy1(0) dy2(0)]

vars = 

values = [c1 c2 c3 c4]

values = 1×4

1.0000 1.0000 2.4495 -2.4495

y1sol = subs(y1sol,vars,values)

y1sol = 

y2sol = subs(y2sol,vars,values)

y2sol = 

%disp('Enter the range for the plot the graph')
%tmin = input('tmin value:');
%tmax = input('tmax value:');
subplot(2,1,1)
ezplot(y1sol,t)
title('The Graph Of y1(t) Vs t')
ylabel('y1(t)')
xlabel('t')
subplot(2,1,2)
ezplot(y2sol,t)
title('The Graph Of y2(t) Vs t')
ylabel('y2(t)')
xlabel('t')

[Thilini Madhushika]

@ Walter Roberson Thank you 😊

[Mohd Aaquib Khan]

@Walter Roberson I modified solution 1 by you for newer version. I am more interested in the ytrans, (Y(s))

I am getting the laplace equation correctly, but if I try to solve for Y, it shows empty solution. Can you please spot the issue for me.

Thanks

syms s t y(t) Y(s)
Dy = diff(y);
D2y = diff(y, 2);
eqn= D2y+Dy+y== (t+1)^3
lteqn=laplace(eqn,t,s)
neweqn=subs(lteqn,[laplace(y(t),t,s),y(0),Dy(0)],[Y,1,0])
ytrans=solve(neweqn,Y)

[Walter Roberson]

syms s t y(t) Y(s) YY
Dy = diff(y);
D2y = diff(y, 2);
eqn= D2y+Dy+y== (t+1)^3
lteqn=laplace(eqn,t,s)
neweqn=subs(lteqn,[laplace(y(t),t,s),y(0),Dy(0)],[YY,1,0])
ytrans=solve(neweqn,YY)

[Mohd Aaquib Khan]

Thanks, So if we use Y(s) it doesnt work and if we use Y only(YY in your code) it works. Quite weird. Please do explain if there is an explaination for this.

[Walter Roberson]

solve() cannot solve for functions such as Y(s)

Answer 4

[Walter Roberson]

You have

 ''D(y)(0)'

Which is invalid syntax. It is the empty string '' immediately followed by D(y)(0) followed by transpose operation.

[Danielle Lawhorn]

what syntax should I replace the invalid syntax with? Thank you for your help!

[Walter Roberson]

Remove the first of the two '' so that you have 'D(y)(0)'