# Thread: Solving partial of ODE to zero

1. ## Solving partial of ODE to zero

Hi guys, I have attached my derivation in the image below. Im not solving the ode, but I need a relationship in this equation. To solve that, I need to make the final part which is

Code:
 [-B/1.101e-17+BN(t)-betaBN(t)]S(t)
to zero, is there anyway that I could make it zero, any theorem that could solve this, I need this very urgently, I am almost solving my own relationship in my system.

Please suggest what I can do. Thank you.

2. Well, as I see it, there are four ways to make that expression zero.

1. $B=0$. Probably not acceptable, I'm guessing.
2. $S(t)=0$. See above.
3. $N(t)$ is a constant, derived as follows:

$-\dfrac{B}{1.101\times 10^{-17}}+B\,N(t)-\beta\,B\,N(t)=0$

$N(t)-\beta\,N(t)=\dfrac{1}{1.101\times 10^{-17}}$

$N(t)=\dfrac{1}{(1.101\times 10^{-17})(1-\beta)}.$

4. Finally, you could have $\beta$ defined as a function of $t$ thus:

$-\dfrac{B}{1.101\times 10^{-17}}+B\,N(t)-\beta\,B\,N(t)=0$

$-\dfrac{B}{1.101\times 10^{-17}}+B\,N(t)=\beta\,B\,N(t)$

$\beta=\dfrac{-\frac{B}{1.101\times 10^{-17}}+B\,N(t)}{B\,N(t)}.$

Since there are no other variables, and all of the variables show up in the equation in a linear fashion, I see no other ways to make the expression zero.

Hope this helps.

3. Hi Ackbeet, thank you very much for the help, I have actually solved it, thanks a lot! May I ask you one more question? 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.

4. Originally Posted by thavamaran
Hi Ackbeet, thank you very much for the help, I have actually solved it, thanks a lot! May I ask you one more question? 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.
The equation can be solved via an integrating factor.

$\displaystyle \frac{dz}{dt}-\left( \frac{-1+\beta}{\tau_n}{\right)z(t)=\Gamma\left( \frac{I(t)}{qv}\right)$

$\displaystyle e^{\int -\left( \frac{-1+\beta}{\tau_n}\right)dt}= e^{ -\left( \frac{-1+\beta}{\tau_n}\right)t}$

$\displaystyle \frac{d}{dt}\left[ e^{ -\left( \frac{-1+\beta}{\tau_n}\right)t} \cdot z(t)\right]=e^{ -\left( \frac{-1+\beta}{\tau_n}\right)t}\Gamma\left( \frac{I(t)}{qv}\right)$

Now just integrate and solve for $z(t)$

5. Hi there, first of all thank you very much for the reply. I found earlier integrating factor is one of the method, but its rather very confusing, like in this case, I dont get why you picked \displaystyle {-1+\beta}{\tau_n} as the integrating factor? Thank you very much for the guidance.

6. Hi mate, thanks again, I found the solution for my earlier question! Thanks a lot, I will get back to you after the derivation. Thanks!

7. For a first-order linear DE in standard form, which is

$y'+P(x)\,y=Q(x),$

the integrating factor is

$\displaystyle e^{\int P(x)\,dx}.$ That's how TheEmptySet picked the integrating factor.