Using ilaplace function i get answer in terms of ilaplace but i need it in terms of a variable t:

This is my code:

clc;

clear;

syms t y(t) s Y

Dy=diff(y,t); D2y=diff(Dy,t);

F=input("Enter coefficients of [y'' y' y]");

a=F(1);b=F(2);c=F(3);

nh=input('Enter the RHS function f(t)');

eqn=a*D2y+b*Dy+c*y-nh;

LTY=laplace(eqn,t,s);

IC=input("enter initial conditions [y0,y0']");

y0=IC(1);dy0=IC(2);

LTY=subs(LTY,laplace(y(t),t,s),Y);

LTY=subs(LTY,y(0),y0);

LTY=subs(LTY,subs(diff(y(t),t),t,0),dy0);

YS=solve(LTY,Y);

y=ilaplace(YS,s,t)

And when i enter the input:

[1 0 1]

3*(1-heaviside(t-4))+(2*t-5)*heaviside(t-4)

[1,0]

I get the output as

y =

2*ilaplace(1/(s^2*exp(4*s) + s^4*exp(4*s)), s, t) - 2*cos(t) + 3

how to get replace ilaplace in the answer with actual answer as function of t?

Answer 1

[Shivansh]

This can be resolved by using simplify function in the penultimate line of your code (line 16 in attached screenshot).

Below is a link to the documentation for the "simplify" function:

https://www.mathworks.com/help/symbolic/simplify.html

After using simplify function you will get output as

This problem can arise when there is there is presence of function which is not integrable in closed form. For example- "exp(-x^2)" is not integrable in closed form. If the expression cannot be simplified further, it will give output in term of ilaplace only.

However In some cases , when "ilaplace" returns a function of "ilaplace" as part of the output, calling "simplify" before passing the function to "ilaplace" leads to a closed-form solution.

This can be understood better by using the following example. If we execute the following line in Matlab.

syms s  
ilaplace((exp(pi*s) + 1)/((exp(pi*s) + s^2*exp(pi*s))*(s^2 + s + 5/4))) 

we will get result as-

ilaplace(exp(pi*s)/((exp(pi*s) + s^2*exp(pi*s))*(s^2 + s + 5/4)), s, t) +
ilaplace(1/((exp(pi*s) + s^2*exp(pi*s))*(s^2 + s + 5/4)), s, t) 

If, however, one factors "exp(pi*s)" out of the numerator and denominator, one will get output as:

syms s 
ilaplace(1/((s^2 + 1)*(s^2 + s + 5/4)) + exp(-pi*s)/((s^2 + 1)*(s^2 + s + 5/4))) 

This returns:

(4*sin(t))/17 - (16*cos(t))/17 + (16*exp(-t/2)*(cos(t) + sin(t)/4))/17 - 
heaviside(t - pi)*((4*sin(t))/17 - (16*cos(t))/17 + (16*exp(pi/2 - t/2)*(cos(t) + sin(t)/4))/17) 

which is a closed-form solution.

Similarly, if one calls:

syms s  
ilaplace(simplify((exp(pi*s) + 1)/((exp(pi*s) + s^2*exp(pi*s))*(s^2 + s + 5/4)))) 

This returns:

(4*sin(t))/17 - (16*cos(t))/17 + (16*exp(-t/2)*(cos(t) + sin(t)/4))/17 -
4*heaviside(t - pi)*(sin(t)/17 - (4*cos(t))/17 + (4*exp(pi/2 - t/2)*(cos(t) + sin(t)/4))/17) 

Which is the same output after factoring out "exp(pi*s)"(after simplification).