How Can I Solve a Unique First-Order ODE Numerically?

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SUMMARY

The discussion focuses on solving a unique first-order ordinary differential equation (ODE) of the form (da/dt)^2 + f(a)*(da/dt) = g(a), where a(t) is the unknown function and f(a) and g(a) are known functions. The user proposes a numerical approach by substituting da/dt with y, transforming the equation into a solvable format. The iterative scheme involves calculating a(i+1) based on the derived values of y and a(i) using a small time step (dt=0.001). The user seeks validation of their reasoning and hints at utilizing the quadratic formula for solving the equation.

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makris
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I would like to solve a problem of the type

(da/dt)^2 + f(a)* (da/dt) = g(a) (1)

a=a(t) unknown function
f(a), g(a) = known functions of a.

This differential equation is a first order ODE but (da/dt)^2 makes it different compared to a typical first order ODEs (at least to my knowledge)

I would like to find a(t) satisfying (1) subject to certain initial conditions (say a(0.1)=2).

I feel that no appropriate analytical solution exists for this type of problem, so I am looking for a numerical method to integrate it.

I am thinking of setting da/dt=y thus having
---------------------------
y^2 + f(a)* y = g(a)
da/dt=y
----------------------------

and then writing da/dt = ( a(i+1) - a(i) ) /dt

so the problem becomes
---------------------------
y^2 + f(a(i))* y = g(a(i)) (2)
a(i+1) = dt*y + a(i) (3)
----------------------------

Now I am thinking of solving (2) for the value of y which corresponds at i=0 and then keep one the two solutions (which one to keep is not very clear ….(or if they are both imaginaray?)) Then with a selected small dt (say dt=0.001) find a(i+1). Then continue the iteration scheme this way.

I know a priori that da/dt is positive and thus a(t) is an increasing function of t.

I would like to have opinion from you whether the previous reasoning is TOTALLY WRONG or not. If it wrong I would appreciate if you just give a hint of how to attack the problem

Thanks
 
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HINT: Use the quadratic formula. :)
 

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