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

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)

[Walter Roberson]

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.)

Answer 1

[Walter Roberson]

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))

[HONG CHENG]

Thank you for your quick reply. I have one more question.

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)

I tried the following codes,

syms x(t) x0
dx = diff(x)
eqn =dx == x(t)+(x(t))^3+sin(100*t)
X = simplify(dsolve(eqn, x(0)==x0))     %boundary condition on x(0)

But I just got

ans =
 
[ empty sym ]

I also tried

X = simplify(dsolve(eqn)

without the boundary condition, but the ans is the same

[Walter Roberson]

syms x(t)
f(t) = sin(100*t)

f(t) = 

dx = diff(x)

dx(t) = 

eqn = dx == x(t)+(x(t))^3 + f(t)

eqn(t) = 

X = simplify(dsolve(eqn))

Warning: Unable to find symbolic solution.

 
X =
 
[ empty sym ]
 

[Walter Roberson]

Most differential equations do not have a solution that can be found in a reasonable amount of time. Most of them have no closed form solution at all.

In such situations you have to resort to numeric solutions.

See the first example of odeFunction() to see a useful work-flow.

[HONG CHENG]

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]

No, odeFunction() and dsolve() are completely useless for difference equations.