Answer to Question #138258 in Differential Equations for Vandita Kamat

Question

(1+yz)dx + x(z-x)dy -(1+xy)dz=0

Answer 1

$\displaystyle\textsf{This is a Pfaffian differential equation in three} \\\textsf{variables and we must verify its integrabilty} \\\textsf{and determine its primitive.}\\ (1 + yz)\mathrm{d}x + (xz − x^2)\mathrm{d}y − (1 + xy)\mathrm{d}z = 0 \\ \textsf{The necessary and sufficient condition}\\\textsf{for iintegrability is} \\ \textbf{X}\cdot curl\textbf{X} = 0 \\ \textbf{X}=(1 + yz,xz − x^2,-1 - xy), \textsf{so that}\\ \nabla \times X = \begin{vmatrix} \textbf{i} &\textbf{j} &\textbf{k} \\ \frac{\partial}{\partial x} &\frac{\partial}{\partial y} &\frac{\partial}{\partial z}\\ 1 + yz &xz − x^2 &-1 - xy \end{vmatrix} = \\ \begin{aligned} &\textbf{i} \left(\frac{\partial}{\partial y}(-1 - xy) - \frac{\partial}{\partial z}(xz - x^2)\right) - \\&\textbf{j} \left(\frac{\partial}{\partial x}(-1 - xy) - \frac{\partial}{\partial z}(1 + yz)\right) + \\&\textbf{k} \left(\frac{\partial}{\partial x}(xz - x^2) - \frac{\partial}{\partial y}(1 + yz)\right) \end{aligned} \\ \begin{aligned} &=\textbf{i}(-x - x) - \textbf{j}(-y - y) + \textbf{k}(z - 2x - z) \\&= -2x\textbf{i} + 2y\textbf{j} - 2x\textbf{k} \end{aligned} \\ \begin{aligned} \therefore &(1 + yz,xz − x^2,-1 -xy) \cdot (-2x, 2y, -2x) \\&= -2x(1 + yz) + 2y(xz - x^2) + 2x(1 + xy) \\&= -2x - 2xyz + 2xyz - 2yx^2 + 2x + 2x^2y = 0 \end{aligned} \\ \textsf{Thus, the given equation is integrable.}\\ \textsf{Solving by Inspection} \\ (1 + yz)\mathrm{d}x + (xz − x^2)\mathrm{d}y − (1 + xy)\mathrm{d}z = 0\\ (1 + yz)\mathrm{d}x + (xz − x^2)\mathrm{d}y − (1 + xy)\mathrm{d}z = 0\\ -(1 + yz)\mathrm{d}x - (xz − x^2)\mathrm{d}y − (1 + xy)\mathrm{d}z = 0\\ (1 + yx)\mathrm{d}z - (1 + yx)\mathrm{d}x - x(z - x) \mathrm{d}y - y(z - x)\mathrm{d}x = 0\\ (1 + yx)\mathrm{d}(z - x) - (x\mathrm{d}y + y\mathrm{d}x)(z - x) = 0\\ (1 + yx)\mathrm{d}(z - x) - \mathrm{d}(1 + xy)(z - x) = 0 \\ \implies \frac{\mathrm{d}(z - x)}{z - x} - \frac{\mathrm{d}(1 + xy)}{1 + xy} = 0\\ \textsf{Integrating both sides, we have}\\ \int\frac{\mathrm{d}(z - x)}{z - x} - \int\frac{\mathrm{d}(1 + xy)}{1 + xy} = \int0\, \mathrm{d}x\\ \ln(z - x) - \ln(1 + xy) = C_1 \\ \ln(z - x) = \ln(A(1 + xy)), \hspace{1cm} \{C_1 = \ln(A)\}\\ \implies z = x + A(1 + xy) \\ \therefore z = x + A (1 + yx)\hspace{0.1cm}\textsf{is a solution to the Pfaffian differential equation}$