Answer to Question #175837 in Differential Equations for Selvi

Question

Solve the pfaffian D.E. y(x+4)(y+z)dx - x(y+3z)dy+2xydz =0

Answer 1

$y(x+4)(y+z)dx-x(y+3z)dy+2xydz=0\\ X=(P,Q,R)\text{ --- vector field}.\\ X\cdot \text{curl } X=0\\ X=(y(x+4)(y+z),-x(y+3z),2xy)\\ \text{curl }X=det \begin{bmatrix} \overline{i} & \overline{j} & \overline{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ P & Q &R \end{bmatrix}=\\ =5x\overline{i}+(2y+yx)\overline{j}+(-9y-7z-2xy-xz)\overline{k}\\ \text{We can check that }X\cdot \text{curl } X=0.\\ \text{Let }x=const.\text{ Then:}\\ -x(y+3z)dy+2xydz=0|:(-x)\\ (y+3z)dy-2ydz=0|:dy\\ y+3z-2yz^{\prime}=0|:2y\neq 0\\ z^{\prime}=\frac{y+3z}{2y}\\ z^{\prime}=\frac{1}{2}+\frac{3}{2}\cdot\frac{z}{y}\\ t(y)=\frac{z}{y};\ z=t\cdot y;\ z^{\prime}=t^{\prime}\cdot y+t.\\ t^{\prime}\cdot y+t=\frac{1}{2}+\frac{3}{2}t\\ t^{\prime}\cdot y-\frac{1}{2}\cdot t-\frac{1}{2}=0\\ \frac{dt}{dy}y-\frac{1}{2}t-\frac{1}{2}=0\\ ydt-\frac{1}{2}tdy-\frac{1}{2}dy=0\\ ydt+(-\frac{1}{2}t-\frac{1}{2})dy=0|:y(-\frac{1}{2}t-\frac{1}{2})\neq 0\\ \frac{dt}{-1/2t-1/2}+\frac{dy}{y}=0\\ -2\frac{dt}{t-1}+\frac{dy}{y}=0|\int\\ -2\ln|t-1|+\ln|y|=\ln C\\ \ln\frac{|y|}{(t-1)^2}=\ln C\\ \frac{y}{(t-1)^2}=C\\ U(x,y,z)=\frac{y}{(z/y-1)^2}=C.\\ \exist \mu:\ \mu Q=\frac{\partial Q}{\partial y}.\\ \frac{\partial Q}{\partial y}=\frac{-z-y}{y(z/y-1)^7}\\ \mu=\frac{-z-y}{y(z/y-1)^7}\cdot \frac{1}{x(y+3z)}=\frac{z+y}{xy(y+3z)(z/y-1)^7}.\\ k=\mu P-\frac{\partial U}{\partial x}.\\ \frac{\partial U}{\partial x}=0\\ k=\mu P\\ k=\frac{(y+z)^2(x+4)}{x(y+3z)(z/y-1)^7}\\ \frac{\partial U}{\partial x}+k=0\\ \int kdx=C\\ U(x,y,z)=\frac{(x+4)(y+z)^2}{x(y+3x)(z/y-1)^7}=C.$