A basic question on first order linear PDEs

Question

I can not figure out how we obtained the ratio:

ِA follow up question:

Where do we use the assumption that $u$ and $v$ are linearly independent ? Thanks a lot

Answer 1

I think this will clarify your doubts.

(I just skipping the previous part)

$$\frac{\partial u}{\partial x}~dx ~+~\frac{\partial u}{\partial y}~dy ~+~\frac{\partial u}{\partial z}~dz ~=~0 \tag1$$ $$\frac{\partial v}{\partial x}~dx ~+~\frac{\partial v}{\partial y}~dy ~+~\frac{\partial v}{\partial z}~dz ~=~0 \tag2$$ By cross-multiplication, $$\frac{dx}{\frac{\partial u}{\partial y}~\frac{\partial v}{\partial z}~-~\frac{\partial u}{\partial z}~\frac{\partial v}{\partial y}}=\frac{dy}{\frac{\partial u}{\partial z}~\frac{\partial v}{\partial x}~-~\frac{\partial u}{\partial x}~\frac{\partial v}{\partial z}}=\frac{dz}{\frac{\partial u}{\partial x}~\frac{\partial v}{\partial y}~-~\frac{\partial u}{\partial y}~\frac{\partial v}{\partial x}}\tag3$$ which can be written in the following ratio form,$$dx:dy:dz=\left(\frac{\partial u}{\partial y}~\frac{\partial v}{\partial z}~-~\frac{\partial u}{\partial z}~\frac{\partial v}{\partial y}\right):\left(\frac{\partial u}{\partial z}~\frac{\partial v}{\partial x}~-~\frac{\partial u}{\partial x}~\frac{\partial v}{\partial z}\right):\left(\frac{\partial u}{\partial x}~\frac{\partial v}{\partial y}~-~\frac{\partial u}{\partial y}~\frac{\partial v}{\partial x}\right)\tag4$$ Again $$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\tag5$$can also written in the following ration form $$dx:dy:dz=p:Q:R\tag6$$ Combining $(4)$ and $(6)$, we have $$dx:dy:dz=\left(\frac{\partial u}{\partial y}~\frac{\partial v}{\partial z}~-~\frac{\partial u}{\partial z}~\frac{\partial v}{\partial y}\right):\left(\frac{\partial u}{\partial z}~\frac{\partial v}{\partial x}~-~\frac{\partial u}{\partial x}~\frac{\partial v}{\partial z}\right):\left(\frac{\partial u}{\partial x}~\frac{\partial v}{\partial y}~-~\frac{\partial u}{\partial y}~\frac{\partial v}{\partial x}\right)=P:Q:R$$