# solution of ordinary differential equations when there is a f(t)

2 Ansichten (letzte 30 Tage)
HONG CHENG am 5 Mär. 2021
Kommentiert: Walter Roberson am 5 Mär. 2021
I do hope anyone can give me some idea to solve these two problems shown in two boxs I can calculte the the solution of x(t) in the following equation
dx(t)/dt = x(t)+(x(t))^3
I can use
dsolve('Dx=1*x+1*x^3')
and I got the answer is
ans =
0
(-exp(2*C8 + 2*t)/(exp(2*C8 + 2*t) - 1))^(1/2)
1i
-1i
I don't know what's C8 and should I just take the (-exp(2*C8 + 2*t)/(exp(2*C8 + 2*t) - 1))^(1/2) as the correct solution?
More important, I don't know how to calculte the solution of x(t) when there is a f(t)
dx(t)/dt = x(t)+(x(t))^3 + f(t)
, where
f(t) = sin(100*t)
##### 1 KommentarKeine anzeigenKeine ausblenden
Walter Roberson am 5 Mär. 2021
I cannot read some of the details of f(t) for the second equation.
Maple and Mathematica both say that there is no closed form solution for the first equation, and no closed form solution for diff(x(t), t) == x(t) + cos(t)^8 + x(t)^3 + 2*sin(5*t)*exp(t) + 1 (which is the best I could estimate for the second equation.)

Melden Sie sich an, um zu kommentieren.

### Akzeptierte Antwort

Walter Roberson am 5 Mär. 2021
I don't know what's C8 and should I just take the (-exp(2*C8 + 2*t)/(exp(2*C8 + 2*t) - 1))^(1/2) as the correct solution?
Yes? No?
C8 represents a constant needed to represent a boundary condition.
syms x(t) x0
dx = diff(x)
dx(t) = eqn = dx == x(t)+(x(t))^3
eqn(t) = X = simplify(dsolve(eqn, x(0)==x0)) %boundary condition on x(0)
X = subs(X,t,0) %crosscheck
ans = Oh dear, that loses the sign. What happens if x0 was negative?
Xneg = dsolve(eqn, x(0)==-2)
Warning: Unable to find symbolic solution.
Xneg = [ empty sym ]
Xpos = simplify(dsolve(eqn, x(0)==2))
Xpos = fplot(Xpos, [0 1]) The larger the boundary condition, the smaller the distance until the singularity. For small enough boundary conditions, the distance to the singularity is approximately -log(sqrt(x0)) -- for boundary conditions of the form 1/N for large enough N, that would be very close to log(sqrt(N))
##### 5 Kommentare4 ältere Kommentare anzeigen4 ältere Kommentare ausblenden
Walter Roberson am 5 Mär. 2021
No, odeFunction() and dsolve() are completely useless for difference equations.

Melden Sie sich an, um zu kommentieren.

### Kategorien

Find more on Equation Solving in Help Center and File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!