Is it possible to solve this differential equation?

  • #1

Main Question or Discussion Point

Hello. I'm solving the second order nonlinear ODE, and I'm not sure that it's possible or not.

Please help.

A_0, B_0, C_0, a, b, c, e, p is all known constant.

dA/dt = -(a+b*C)*B
dB/dt = -p*c*A
dC/dt = -(1-p)*e*A

then I want to get a solution like A(t) = function of (A_0, B_0, C_0, a, b, c, e, p)
Can it be possible?

please help :(.
 

Answers and Replies

  • #2
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Find C in terms of A and B from the first equation and then put it in the third equation. This gives you an equation involving only A and B. Then you can write one of them in terms of the other and put it in the second equation. Now you have an ordinary differential equation involving only one unknown function.
 
  • #3
pasmith
Homework Helper
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410
Hello. I'm solving the second order nonlinear ODE, and I'm not sure that it's possible or not.

Please help.

A_0, B_0, C_0, a, b, c, e, p is all known constant.

dA/dt = -(a+b*C)*B
dB/dt = -p*c*A
dC/dt = -(1-p)*e*A

then I want to get a solution like A(t) = function of (A_0, B_0, C_0, a, b, c, e, p)
Can it be possible?

please help :(.
If [itex]pc = 0[/itex] then [itex]B[/itex] is constant, and the system for [itex]A[/itex] and [itex]C[/itex] is then linear, so in principle you can solve it analytically. Similarly if [itex]e(1 -p) = 0[/itex] then [itex]C[/itex] is constant, and the system for [itex]A[/itex] and [itex]B[/itex] is then linear, so again in principle you can solve it analytically.

Aside from those cases, the system is non-linear, so probably cannot be solved analytically and you may have to rely on numerical methods. The quantity [itex]X = e(1-p)B - pcC[/itex] is conserved so the trajectory is confined to a plane of constant [itex]X[/itex]. Thus you can eliminate [itex]C[/itex] and consider the second-order system [tex]
\dot A = -\left(a + \frac{b}{pc}(e(1-p)B - X)\right)B \\
\dot B = -pcA.[/tex]

EDIT: We have [itex]\ddot B = -pc\dot A = (pca + b(e(1-p)B - X)B[/itex] which is a function of [itex]B[/itex]; therefore one can multiply both sides by [itex]\dot B[/itex] to obtain [tex]\frac{d}{dt}\left( \tfrac12 (\dot B)^2 - \tfrac13 be(1-p)B^3 + \tfrac12 bXB^2 - pcaB \right) = 0[/tex] which you can then integrate easily to obtain an equation for [itex]\dot B[/itex] in terms of [itex]B[/itex] (be careful with the choice of sign when taking the root). Unfortunately you probably can't then solve analytically for [itex]B[/itex], but if you could you would then have [tex]
A(t) = A_0 - \int_0^t \left(a + \frac{b}{pc}(e(1-p)B(s) - X)\right)B(s)\,dt[/tex] but you probably can't do that integral analytically even if you knew [itex]B[/itex].
 
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