# Thread: Second Order Differential Equation

1. ## Second 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.

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

Thank you very much.

2. Originally Posted by jen81
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.

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

Thank you very much.
$\frac{d^2Q}{dn^2} - A(C - n)\frac{dQ}{dn} + \left(A - \frac{B}{n}\right)Q = 0$.

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

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

Solve this Quadratic equation for $m$.

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.

3. Hi,

Thanks ProveIt.
The solution to the equation above if assuming B = 0 is
$
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.

4. 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:

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

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

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

which then gets it to:

$v''+Iv=0$

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

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