# How to solve second diffrential equation that look like this ?

조회 수: 1(최근 30일)
Nora Khaled 2 Oct 2018
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{:});
sys_sym = [sys{~chars}];
Error in dsolve (line 189)
Error in t1 (line 17)
Y = dsolve(sodeq,diff(y(t))==0,y(t)==3);
which I do not know what it means?

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### 답변 수 (2)

ANKUR KUMAR 3 Oct 2018

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Nora Khaled 5 Oct 2018
Thank you!

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Walter Roberson 4 Oct 2018
Walter Roberson 님이 편집함. 5 Oct 2018
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.

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Nora Khaled 5 Oct 2018