Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$$-A_g{d^2g\over dx^2} + B_g{dg\over dx} = C - Mgh \tag 1$$ $$-A_h{d^2h\over dx^2} + B_h{dh\over dx} = C - Mgh \tag 2$$

where, $A_g, B_g, A_h, B_h, C$ and $M$ are constants. And, $g$ and $h$ are dependent variables and $x$ is the independent variable.

I have tried to solve the problem in following way:

(1) - (2) $\Rightarrow -A_gD^2g + A_h D^2h + B_g Dg - B_hDh = 0$ [where, $D \equiv {d\over dx} $]

$\Rightarrow h = {B_gDg - A_gD^2g\over B_hD - A_hD^2} = {B_gg - A_gDg\over B_h - A_hD}$ [I am confused about this step. Can we express $h$ in this way?]

Now putting $h$ in (1) we can solve (1).

Is my procedure okay? And, is there any other way to solve the problem more easily and/or correctly?

share|cite|improve this question
Until 1)-2) it is ok, but then you "remove" the diff. operator $D^2$ from $h$ and this is not correct – Avitus Nov 11 '13 at 18:33
up vote 0 down vote accepted

After subtracting (2) from (1) and setting $z=g-h$ we will have a second order homogeneous equation in $x$ and $z$: $-A_g z''+B_g z'=0$. Solving this equation we get $$z=c_1+c_2 \displaystyle e^{-\frac{B_g}{A_g}x}.$$ So $g= h+c_1+c_2 \displaystyle e^{-\frac{B_g}{A_g}x}$ and hence substituting this into the second one you will have a second order equation for $h$, but unfortunetly, it is nonlinear.

share|cite|improve this answer
Thank you very much for your answer. But in my case $A_g\neq A_h$ and $B_g \neq B_h$. It would be great if you please help me for this case also. – crazy particle Nov 19 '13 at 20:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.