Solving Forward Equation I've currently started reading 'Lectures on Partial Differential Equations' by Faris.
Page 44 he states the following forward equation:
$$J=a(y)p-\frac{1}{2}\frac{\partial \sigma(y)^2p}{\partial y} = 0$$
I understand how to solve this and obtain the following:
$$I(y) = \int_{y_0}^{y}\frac{2}{\sigma(z)^2}a(z)dz$$
i.e. integrate once and separate the variables. He then states solving for $p$ produces:
$$p(y)=C\frac{2}{\sigma(y)^2}exp(I(y))$$
I don't understand where the $\frac{2}{\sigma(y)^2}$ comes from. Can someone guide me in the right direction please?
 A: It's been four days and you've managed to solve your own question. I'll now share it with you.
Note $J=0$ and $\sigma(y)$ is a function on of variable (i.e. there is no time variable).


*

*Rearrange the Forward Equation and use standard differential operators:
$$a(y)p = \frac{1}{2}\frac{d(\sigma(y)^2p)}{dy}$$

*Multiply through by 2 and use the product rule:
$$2a(y)p=\frac{d(\sigma(y)^2)}{dy}p+\frac{d(p)}{dy}\sigma(y)^2$$

*Now divide through by p and divide through by $\sigma(y)^2$:
$$\frac{2a(y)}{\sigma(y)^2}=\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}+\frac{1}{p}\frac{d(p)}{dy}$$

*Use an integration 'trick':
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{1}{\sigma(y)^2}\frac{d(\sigma(y)^2)}{dy}dy+\int_{}{}\frac{1}{p}\frac{d(p)}{dy}dy$$

*Simplify!
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\int_{}{}\frac{d(\sigma(y)^2)}{\sigma(y)^2}+\int_{}{}\frac{d(p)}{p}$$

*Perform rhs integration:
$$\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy=\ln \sigma(y)^2 +\ln P +\ln C$$

*Exponential and rearrange:
$$p(y) = C\frac{1}{\sigma(y)^2}exp(\int_{}{}\frac{2a(y)}{\sigma(y)^2}dy)$$
Note:
I am not sure where Mr Faris obtained the '2' in the final solution. Perhaps someone more knowledgable will be able to explain. I believe it is a mistake and the constant of integration should be correct.
