Hi,

I have a second order differential equation as shown below. I have no idea how to start and solve the problem. Appreciate if anyone could help me on this.

$\displaystyle

(A - B/n)Q - A(C-n)dQ/dn + d2Q/dn2 = 0

$

Thank you very much.

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- Nov 1st 2009, 04:17 PMjen81Second Order Differential Equation
Hi,

I have a second order differential equation as shown below. I have no idea how to start and solve the problem. Appreciate if anyone could help me on this.

$\displaystyle

(A - B/n)Q - A(C-n)dQ/dn + d2Q/dn2 = 0

$

Thank you very much. - Nov 1st 2009, 04:40 PMProve It
$\displaystyle \frac{d^2Q}{dn^2} - A(C - n)\frac{dQ}{dn} + \left(A - \frac{B}{n}\right)Q = 0$.

I'm assuming that this is a second order linear CONSTANT COEFFICIENT ODE, so write the characteristic equation:

$\displaystyle m^2 - A(C - n)m + \left(A - \frac{B}{n}\right)Q = 0$.

Solve this Quadratic equation for $\displaystyle m$.

You will end up with either:

1. Two real distinct solutions

2. Two real repeated solutions

3. Complex conjugates.

The solution to the ODE depends on the solutions of the characteristic equation. - Nov 2nd 2009, 12:58 AMjen81
Hi,

Thanks ProveIt.

The solution to the equation above if assuming B = 0 is

$\displaystyle

C1\exp[-A/2(n-C)^2]\int \exp[A/2(n'-C)^2]dn'); C1 = constant

$

I tried the method as suggested but I couldn't get to this solution.

Thank you very much. - Nov 2nd 2009, 02:08 PMshawsend
The differentiation is with respect to n, albeit non-standard notation, so it's a DE with variable coefficients. And if I let B=0 and just change n to x to make it more standard looking, it's then:

$\displaystyle Q''-A(C-x)Q'+AQ=0$

So other than power series, we could get it into normal form letting:

$\displaystyle Q=v\,\text{exp}\left\{1/2\int A(C-x) dx\right\}$

which then gets it to:

$\displaystyle v''+Iv=0$

with: $\displaystyle I=1/2A+1/4 A^2(C-x)^2$

but that's as far as I can take it.