How Do You Solve a Nonlinear Differential Equation Involving Exponential Terms?

  • Thread starter Thread starter CJDW
  • Start date Start date
  • Tags Tags
    Nonlinear
CJDW
Messages
3
Reaction score
0
Nonlinear DE (with e^t) ?

Good day forum,

I have this wonderful DE :

dx/dt = [a - f '(t)]x + (b + d(c^t))(x^2) - 1

with,
t \in [s,T]
x(T) = 0

a, b, d & c are constants.
f(t) = g + h(k^t) , where g, h & k are constants (but I think specifying this is of no importance)

My knowledge of non-linear equations is very limited and would sincerely appreciate any help whatsoever.

CJDW
 
Physics news on Phys.org


That looks like a Riccati equation:

\frac{dx}{dt}=\left(a-f'(t)\right)x+(b+dc^t)x^2-1

\frac{dx}{dt}+Q(t)x+R(t)x^2=P(t)=-1

and using the standard transformation for a Riccati equation, obtain a second-order (linear) DE:

Ru''-(R'-QR)u'-PR^2u=0

Now, you can then put the equation in it's Normal form by letting:

u=v\text{exp}\left(-1/2\int P dt\right)

in order to remove the term involving the first derivative. Yeah, I know this ain't easy. I'm getting this right out of "Intermediate Differential Equations" by Rainville. We then obtain the equation:

v''+Iv=0

where:

I=Q-1/2 P'-1/2 P^2

and if I just happens to be a constant, that equation can be easily solved.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top