Help with Laplace ODE

Help solving this laplace ode with intial condition, my a and b values are respectively. I have tried solving this Laplace ode without using my a and b values. below is my code.

Any help is massively appreciated! Thank you

syms t y(t) Y(s) Dy0 a b y0
Dy1 = diff(y,t,1); 
eqn = laplace(Dy1== a*sinh(a*t));
eqn = subs(eqn, {laplace(y(t), t, s), subs(diff(y(t), t), t, 0), y(0)},{Y(s), 0,0})

cond= y(0)==22;

Ys(s)=  simplify(eqn/Y);
yt = ilaplace(Ys,s,t)

Answer 1

[Star Strider]

Try this —

syms t y(t) Y(s) Dy0 a b y0
Dy1 = diff(y,t,1); 
eqn = laplace(Dy1== a*sinh(a*t))

eqn = 

eqn = subs(eqn, {laplace(y(t), t, s), subs(diff(y(t), t), t, 0), y(0)},{Y(s), 0, 22})   % Substitute 'y(0)' Here

eqn = 

% cond= y(0)==22;
eqn = isolate(eqn, Y)

eqn = 

y(t,a) = ilaplace(rhs(eqn))

y(t, a) = 

figure
fplot(y(t,34), [-0.1  0.1])                                                             % Substitute 'a' Here
grid

.

[gurjeevan singh]

THANK YOU SO MUCH !

Perfect!

[Star Strider]

As always, my pleasure!

[gurjeevan singh]

How would I adapt the eqn formula to add another variable, for example:

where; is the added term.

I tried changing the symbolic functions, but I am missing a term in the second eqn formula (subs). What do I need to do to solve this different ODE laplace ?, using the above solution?

Once again, any help massively appreciated. Thank you for answering my previous question

[Star Strider]

I would do something like this (the code is essentially the same) —

syms t y(t) Y(s) Dy0 a b y0 beta 
Dy1 = diff(y,t,1); 
eqn = laplace(Dy1 + beta*y == a*sinh(a*t))

eqn = 

eqn = subs(eqn, {laplace(y(t), t, s), subs(diff(y(t), t), t, 0), y(0)},{Y(s), 0, 22})   % Substitute 'y(0)' Here

eqn = 

eqn = isolate(eqn, Y)

eqn = 

y(t,a,beta) = ilaplace(rhs(eqn))

y(t, a, beta) = 

You may be able to simplify that further if you wish, and perhaps using partfrac before ilaplace.

To plot it with fplot, proivide values for both scalar parameters.

[gurjeevan singh]

Once again, Thank you so much!

This is helping massively ! Made my day !

[Star Strider]

As always, my pleasure!

[gurjeevan singh]

Sorry to bother again, now, I am trying to solve a laplace 2nd order equation:

with conditions , .

To solve this I have slighlty adapted the code, however Im not sure where to put my new limit of Dy(0) (shown in code)

syms t y(t) Y(s) Dy0 a b y0

Dy1 = diff(y,t,1);
Dy2 = diff(y,t,2);

eqn = laplace(Dy2-10*Dy1 +9*y==5*t)

%%%
eqn = subs(eqn, {laplace(y(t), t, s), subs(diff(y(t), t), t, 0)},{Y(s), Dy0});
cond = [y(0)==-1, Dy1(0)==2];
%%%

eqn=isolate(eqn, Y)
y(t) = ilaplace(rhs(eqn))

I need to relate my condition back into the code, how would I do this ?

This code actually solves the 2nd order ODE, but at the end I have to plug in the values

for and . Below shows an attempt at finding solving the 2nd order by introducing another subs term. However, this does not work, because I need to define a condition for Dy(0) and theres a paranthesis error in the t,t,),t.

eqn= subs(eqn, {laplace(y(t), t, s), subs (diff(y(t),t,t),t ,0), subs(diff(y(t), t), t, 0), y(0), Dy1(0)},{Y(s), 0, -1});

Also, before I ask again, to solve a 3rd order laplace ODE, I would have to repeat the same added procedural steps for solving a 2nd order ODE?

Any help massively appreciated!

[Star Strider]

I believe this does what you want —

syms t y(t) Y(s) Dy0 a b y0
Dy1 = diff(y,t,1);
Dy2 = diff(y,t,2);
eqn = laplace(Dy2-10*Dy1 +9*y==5*t)

eqn = 

eqn = subs(eqn, {laplace(y(t), t, s), subs(diff(y(t), t), t, 0), y(0)},{Y(s), 2, -1})

eqn = 

%%%
eqn=isolate(eqn, Y)

eqn = 

y(t) = ilaplace(rhs(eqn))

y(t) = 

figure
fplot(y, [0 1])
grid
xlabel('t')
ylabel('y(t)')

Check to be sure the substitutions are correct.

[gurjeevan singh]

Thank you so much x10. Made my day again !

massively appreciated, words cant express it !

[Star Strider]

As always, my pleasure!

(A Vote would be appreciated!)