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$$a \equiv a(\sigma,t)$$ $$b \equiv b(\sigma,t)$$ $$ \frac{1}{2}\frac{d^2}{d\sigma^2}(b^2p) - \frac{d}{d\sigma}(ap) = 0$$

Show that the following satisfies the above differential equation: $$p(\sigma) = \frac{A}{b^2}e^{\int^\sigma \frac{2a}{b^2} d\sigma'}$$

Considering the boundary conditions that as $\sigma \to \infty$, $p \to 0$ and $\frac{dp}{d\sigma} \to 0$

Attempted solution: the line of attack is to differentiate with respect to $\sigma$ given what we have for $p$.

So, starting with the first order differential:

$$\frac{d}{d\sigma}(ap) = a.p' + a'p$$

Exuse the mixed notation style. Given the boundary conditions in the question don't both of these terms become 0?

Also, is there a way to calculate the value of A given the boundary conditon. I asked a related question here but we ended up finding the solution to the DE, instead of showing that the given solution satisfies the DE. Hence this new question.

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Given the boundary conditions in the question don't both of these terms become 0?

The boundary conditions apply on the boundary. The verification of equation is done inside the domain. These are unrelated things.

Actually, you don't need to find the derivative of $ap$. Begin from the other end: take the derivative of $$ b^2p = A\exp\left( \int^\sigma \frac{2a}{b^2} d\sigma' \right) \tag1$$ with respect to $\sigma$: $$ \frac{\partial}{\partial\sigma}(b^2p) = A\,\frac{2a}{b^2}\exp\left( \int^\sigma \frac{2a}{b^2} d\sigma' \right) = 2ap \tag2$$ Thus, the equation becomes $$ \frac{1}{2}\frac{\partial}{\partial\sigma}(2ap) - \frac{\partial}{\partial\sigma}(ap) = 0 $$ which is obviously true.

I don't see how one could get $A$ from the boundary conditions you have, unless $A$ is zero. Notice that if $p$ is a function that satisfies all of the stated conditions, then $2p$, $3p$, etc also satisfy them.

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