Solving ordinary differential equation

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Hi guys! Attached is the differential equation that I want to solve both numerically and analytically, numerically done.

But analytically what method could I use? There is so many methods in differential equations. Please advice on this. Thank you very much
 

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HallsofIvy
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Your equation is
[tex]\frac{dZ}{dt}+ \Gamma\frac{I(t)}{qV}= Z(t)\left(\frac{-1+ \beta}{\tau_n}\right)[/tex]


Are [itex]\Gamma[/itex], q, V, [itex]\beta[/itex], and [itex]\tau_n[/itex] all constants? If so that is a "linear non-homogeneous first order differential equation" and there are a number of ways of solving it. For simplicity, I am going to write A for the constant
[tex]\frac{-1+ \beta}{\tau_n}[/tex]
and B for
[tex]\frac{\Gamma}{qV}[/tex]

Now the equation is
[tex]\frac{dZ}{dt}+ BI(t)= AZ[/tex]
or, equivalently,
[tex]\frac{dZ}{dt}- AZ= -BI(t)[/tex]

The simplest method is to find an "integrating factor". That is, find a function, u(t), so that the left side of that equation, multiplied by u(t), is simply
[tex]\frac{d(uZ)}{dt}[/tex].

Of course, by the chain rule,
[tex]\frac{d(uZ)}{dt}= u\frac{dZ}{dt}+ \frac{du}{dt}Z[/tex]
while just multiplying by u would make the left side
[tex]u\frac{dZ}{dt}- AuZ[/tex]

Comparing those, it is easy to see that we need to have
[tex]\frac{du}{dt}= -Au[/itex]
which is the same as
[tex]\frac{du}{u}= -Adt[/itex]

Integrating both sides gives ln|u|= -At+ C or
[tex]U(t)= e^{-At+ C}[/itex]

Since we only need one function, take C= 0 so that [itex]u(t)= e^{-At}[/itex].

That is, multiplying the entire equation by [itex]e^{-At}[/itex] we get
[tex]e^{-At}\frac{dZ}{dt}- Ae^{-at}Z= \frac{d(e^{-at}Z)}{dt}= -Be^{-At}I(t)[/itex]
and so
[tex]e^{-At}Z(t)= -B\int^t e^{-Ax}I(x)dx+ C[/tex]
or
[tex]Z(t)= -Be^{At}\int^t e^{-Ax}I(x)dx+ Ce^{At}[/tex]
where I have changed the "dummy" variable in the integral to "x" so as not to confuse it with the actual variable t. In particular, the two exponentials, one inside the integral and one outside, do not cancel. The upper limit on the integral is to indicate that the variable becomes "t" after the integration and the "C" is the constant of integration.

Of course, exactly what that integral is depends upon the function I(t) and, of course, you want to replace A and B with the original expressions.
 
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  • #3
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Hi HallsofIvy, thank you very much and you are simply great to offer such help. But I still have trouble with solving the part as attached.

Im having trouble solving [tex]e^{-At}\frac{dZ}{dt}- Ae^{-at}Z= \frac{d(e^{-at}Z)}{dt}= -Be^{-At}I(t)[/tex] where im having trouble with integration by parts cause of the variables, its confusing me too much, can you explain to me a bit further on this, im extremely sorry, but i have been trying to solve this since yesterday, deriving and reaching again to different types of false answers.

By the way, the equation is [tex]\frac{dZ}{dt}=\Gamma\frac{I(t)}{qV}+ Z(t)\left(\frac{-1+ \beta}{\tau_n}\right)[/tex] and for integrating factor it should be:

[tex]\frac{dZ}{dt}-Z(t)\left(\frac{-1+ \beta}{\tau_n}\right)=\Gamma\frac{I(t)}{qV}[/tex] .

thank you a millions times!!
 

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