# Solution of Apostol calculus v 2 exercise 11.22 n. 9

I am looking at the solution to this exercise that says

Suppose the differential equation $$P(x,y)dx+Q(x,y)dy=0$$ has an integrating factor $$\mu(x,y)$$ which leads to a one-parameter family of solutions of the form $$\phi(x,y)=C$$. If the slope of the curve $$\phi(x,y)=C$$ at $$(x,y)$$ is $$tan\theta$$, the unit normal vector $$\pmb{n}$$ is taken to mean $$\pmb{n}=sin\theta\pmb{i}-cos\theta\pmb{j}.$$ There is a scalar field $$g(x,y)$$ such that the normal derivative of $$\phi$$ is given by the formula $$\frac{\partial\phi}{\partial n}=\mu(x,y)g(x,y),$$ where $$\frac{\partial\phi}{\partial n}=\nabla\phi\cdot\pmb{n}$$. Find an explicit formula for $$g(x,y)$$ in terms of $$P(x,y)$$ and $$Q(x,y)$$.

The solution of the exercise (I cheated, I know) reported at the end of the book is

$$g(x,y)=\pm\sqrt{P^2(x,y)+Q^2(x,y)}$$

What I observed to justify this is that if the equation $$P(x,y)dx+Q(x,y)dy=0$$ has an integrating factor $$\mu(x,y)$$, it means that the equation $$\mu(x,y)P(x,y)dx+\mu(x,y)Q(x,y)dy=0$$ is exact, i.e. the field $$\mu(x,y)P(x,y)\pmb{i}+\mu(x,y)Q(x,y)\pmb{j}$$ is the gradient of the field $$\phi(x,y)$$. I then observe that being the gradient always orthogonal to level curves, it is always aligned with the unit outer normal $$\pmb{n}$$, hence the scalar product $$\nabla\phi\cdot\pmb{n}$$ is the magnitude of the gradient itself, $$\sqrt{\mu^2(x,y)P^2(x,y)+\mu(x,y)^2Q^2(x,y)}$$, and the solution derives from the fact that an integrating factor is always non zero when needed.

Does this logic make sense? I don't understand how the definition of unit normal vector in terms of $$\theta$$ could be used... maybe in a more rigorous proof which I fail to grasp. Also, is the solution of this exercise used as a formula to solve a particular class of differential equations?

Thanks

• when you say, “the magnitude of the gradient [of $\phi$] itself,” do you mean $\mu(x,y)\sqrt{P^2(x,y)+Q^2(x,y)}$? Commented Mar 14, 2020 at 20:42
• you are right, let me edit. what do you think of this approach? Commented Mar 14, 2020 at 20:47
• Your approach looks good. Commented Mar 14, 2020 at 20:48

Your approach looks good to me. I solved it a bit more explicitly using $$\theta$$.
If $$\mu (P\,dx + Q\,dy)=0$$ is exact, there exists $$\phi$$ such that $$\nabla \phi = \mu P \mathbf{i} + \mu Q \mathbf{j}$$. So $$\frac{\partial \phi}{\partial x} = \mu P \qquad \frac{\partial \phi}{\partial y} = \mu Q$$ Along level curves $$\phi = C$$, the slope of the tangent line to $$(x,y)$$ is $$\frac{dy}{dx} = - \frac{\partial \phi/\partial x}{\partial \phi/\partial y} = - \frac{P}{Q}$$ Setting $$-\frac{P}{Q} = \tan\theta = \frac{\sin\theta}{\cos\theta}$$, we see a solution with $$\sin\theta = -\frac{P}{\sqrt{P^2+Q^2}}$$ and $$\cos\theta = \frac{Q}{\sqrt{P^2+Q^2}}$$. Then $$\mathbf{n} = - \frac{P}{\sqrt{P^2+Q^2}} \mathbf{i} - \frac{Q}{\sqrt{P^2+Q^2}} \mathbf{j}$$ is a unit normal to the same level curve at $$(x,y)$$. It follows that $$\frac{\partial \phi}{\partial n} = \nabla \phi \cdot \mathbf{n} = \left(\mu P \mathbf{i} + \mu Q \mathbf{j}\right) \cdot \left(- \frac{P}{\sqrt{P^2+Q^2}} \mathbf{i} - \frac{Q}{\sqrt{P^2+Q^2}} \mathbf{j}\right) = -\frac{\mu(P^2+Q^2)}{\sqrt{P^2+Q^2}} = -\mu\sqrt{P^2+Q^2}$$