SOLVE LINEAR DIFERENTIAL EQUATION WITH ONE UNKNOWN

조회 수: 2 (최근 30일)
Tomás Yamamoto
Tomás Yamamoto 2021년 3월 13일
댓글: Walter Roberson 2021년 3월 14일
I have the system of the image, but i want to solve the exatly system but insted of a fixed number where the four goes, i would like to find the range of solutions but with the unknown where the four goes.
I would usually solve this with pencil and paper but my system is 6x6, Thanks in Advance.
AUTHOR edit:
Sorry i think my explanation was not clear, what i would like to solve this ->
eqn1 = du == 3*u + X*v;
eqn2 = dv == -X*u + 3*v;
where X is a parameter, and i would like to find the family of solution's in function of my parameter X.

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Walter Roberson
Walter Roberson 2021년 3월 13일
syms u(t) v(t)
du = diff(u,t);
dv = diff(v,t);
eqn1 = du == 3*u + 4*v;
eqn2 = dv == -4*u + 3*v;
sol = dsolve([eqn1, eqn2])
sol = struct with fields:
v: [1×1 sym] u: [1×1 sym]
sol.u
ans = 
sol.v
ans = 
  댓글 수: 2
Tomás Yamamoto
Tomás Yamamoto 2021년 3월 14일
Sorry i think my explanation was not clear, what i would like to solve this ->
eqn1 = du == 3*u + X*v;
eqn2 = dv == -X*u + 3*v;
where X is a parameter, and i would like to find the family of solution's in function of my parameter X.
Walter Roberson
Walter Roberson 2021년 3월 14일
syms u(t) v(t) X
du = diff(u,t);
dv = diff(v,t);
eqn1 = du == 3*u + X*v;
eqn2 = dv == -X*u + 3*v;
sol = dsolve([eqn1, eqn2])
sol = struct with fields:
v: [1×1 sym] u: [1×1 sym]
simplify(sol.u)
ans = 
simplify(sol.v)
ans = 

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