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

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The discussion centers on solving a nonlinear differential equation (DE) of the form dx/dt = [a - f '(t)]x + (b + d(c^t))(x^2) - 1, which is identified as a Riccati equation. The transformation to a second-order linear DE is outlined, leading to the equation v'' + Iv = 0, where I = Q - 1/2 P' - 1/2 P^2. The solution process involves using the standard transformation for Riccati equations and references "Intermediate Differential Equations" by Rainville for further guidance.

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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 [tex]\in[/tex] [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
 
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That looks like a Riccati equation:

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

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

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

[tex]Ru''-(R'-QR)u'-PR^2u=0[/tex]

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

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

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:

[tex]v''+Iv=0[/tex]

where:

[tex]I=Q-1/2 P'-1/2 P^2[/tex]

and if I just happens to be a constant, that equation can be easily solved.
 

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