# Show that solution satisfies the differential equation

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

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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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.