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

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HONG CHENG
HONG CHENG 2021-3-5
评论： Walter Roberson 2021-3-5
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 个评论
Walter Roberson
Walter Roberson 2021-3-5
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.)

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Walter Roberson
Walter Roberson 2021-3-5
Ran in:
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))
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HONG CHENG
HONG CHENG 2021-3-5
Thank you, Sir.
If our t are discrete values, such as t = 1:1:100, is it possible to use odefunction to get the solution?
Walter Roberson
Walter Roberson 2021-3-5
No, odeFunction() and dsolve() are completely useless for difference equations.

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