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

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HONG CHENG 2021-3-5

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评论： Walter Roberson 2021-3-5

采纳的回答： Walter Roberson

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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)

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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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Answer 1

Walter Roberson 2021-3-5

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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))