I've tried many times to reach the solution of a first order differential equation (of the last equa

Cobie Sadler 2022-02-16 Answered
I've tried many times to reach the solution of a first order differential equation (of the last equation) but unfortunately I couldn't. Could you please help me to know how did he get this solution.
The equation is:
dXdt=kf(kf+kb)X
The author assumed that
X=0 at t=0
Then, he said in his first publication the solution is given as:
X=kf(kf+kb){1exp[(kf+kb)t]}
I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.
In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:
X=kf(kf+kb){1exp[(kf+kb)t]}+X0 exp[(kf+kb)t] and he said X0 was taken to be zero in the first publication.
Can any of you help how did the author get this solution please?
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Expert Answer

ayulyndji
Answered 2022-02-17 Author has 3 answers
It is basically an equation X+αX=β with α=kf+fb and β=kf constants, so you solve the homogeneous equation and get the solution for the unhomogeneous from there...
In this case, we can use an ansatz X=a+bect which leads to:
X=c(bect)=c(Xa)=cXca.
Identifying the terms, we have c=α and ca=β. Lastly, the initial condition X0=a+b gives you b.
Substituting a,b,c with the correct values in X=a+bect gives, after slight manipulations, the answer you are seeking, in terms of α, β, X0.
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Radich606
Answered 2022-02-18 Author has 6 answers
It looks like he changed the initial condition
X(t=0)=0  X0
this means when solving your ode you get this additional term.
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