How to solve second diffrential equation that look like this ?

I am trying to solve this equation using matlab and couple others. but I do not know why it does not work.

d2y(t)/t + 2*dy(t)/t +y(t) = d2x(t)

I use this code:

t = (0:0.01:20)';
unitstep = t>=0;
syms y(t) x(t);
x(t)= (cos(t)+sin(2*t))*unitstep; 
%second order diffrintial equation 
sodeq = diff(y(t),2) + 2*diff(y(t)) + y(t) == diff(x(t),2);
Y = dsolve(sodeq,diff(y(t))==0,y(t)==3);

Error msg:

Error using sym/cat>checkDimensions (line 74)
CAT arguments dimensions are not consistent.

Error in sym/cat>catMany (line 37)
[resz, ranges] = checkDimensions(sz,dim);

Error in sym/cat (line 26)
    ySym = catMany(dim, strs);

Error in sym/horzcat (line 19)
    ySym = cat(2,args{:});

Error in dsolve>mupadDsolve (line 310)
sys_sym = [sys{~chars}];

Error in dsolve (line 189)
sol = mupadDsolve(args, options);

Error in t1 (line 17)
Y = dsolve(sodeq,diff(y(t))==0,y(t)==3);

which I do not know what it means?

Answer 1

[ANKUR KUMAR]

Hi Nora Khaled, refer these links:

https://www.mathworks.com/help/symbolic/solve-a-single-differential-equation.html

https://in.mathworks.com/help/symbolic/solve-differential-equation-numerically-1.html

https://in.mathworks.com/help/symbolic/solve-a-single-differential-equation.html#mw_70f735e2-7aa7-43a2-97e8-a1ef36640a04

[Nora Khaled]

Thank you!

Answer 2

[Walter Roberson]

You defined

t = (0:0.01:20)';
unitstep = t>=0;

so after that, unitstep is a vector of length 2001, not a formula or function.

Then you do

syms y(t) x(t);

That is equivalent to

t = sym('t');
y = symfun( sym('y(t)'), t);
x = symfun( sym('x(t)'), t);

Notice it overwrites t, so afterwards t refers to a symbolic scalar, not a numeric vector.

Then you have

x(t)= (cos(t)+sin(2*t))*unitstep;

which is equivalent to

x = symfun( (cos(t)+sin(2*t))*unitstep, t);

With t now being symbolic scalar, the (cos(t)+sin(2*t)) becomes a symbolic expression. But remember that unitstep is a numeric vector, not a formula or function, so you are multiplying the (cos(t)+sin(2*t)) by that numeric vector of length 2001, giving a symbolic vector of length 2001 . So for any given t, x(t) will return a vector of length 2001.

sodeq = diff(y(t),2) + 2*diff(y(t)) + y(t) == diff(x(t),2);

y(t) is symbolic so there is no problem taking diff() of it, so the left side of that resolves to a symbolic expression. The right hand side is taking the derivative of that vector of length 2001, and so is going to give you a vector of length 2001, so sodeq is going to end up creating a vector of length 2001 of comparisons.

Y = dsolve(sodeq,diff(y(t))==0,y(t)==3);

the internal code constructing the system of equations is not expecting vectors of that size.

... In other words, you should change

t = (0:0.01:20)';
unitstep = t>=0;

to

syms t
unitstep(t) = piecewise(t>=0, 1, 0);
syms y(t) x(t)
x(t)= (cos(t)+sin(2*t))*unitstep; 
sodeq = diff(y(t),2) + 2*diff(y(t)) + y(t) == diff(x(t),2);
Y = dsolve(sodeq,diff(y(t))==0,y(t)==3);

This will promptly fail complaining about the number of equations being wrong compared to the number of variables.

To get further, you need to recognize that y(t)==3 is a definition for y(t) that can be substituted into sodeq and diff(y(t))==0 . Substituting into sodeq gives you

3 == piecewise(0 < t, - 4*sin(2*t) - cos(t), 0)

This is independent of y, but has the potential to solve for t:

exact_t = solve(subs(sodeq,y,3));
approx_t = double(exact_t);

So the solution of the system of equations is y(t) == 3 (which you gave) and t is about 11.53 to solve the x portion.

[Nora Khaled]

Thank you so much for your answer!