Solving the First Order PDE:$$2x z_x + 3y z_y = x + y$$
for $c_1$ I get: $$ c_1 = \frac{x^3}{y^2} $$
I do not know how to go about solving for $c_2$
The correct general solution should be:
$$z = \frac{x}{2} + \frac{y}{3} + f\bigg(\frac{x^3}{y^2}\bigg)$$
How would one come to this calculation?
I would really appreciate your help
$$2x z_x + 3y z_y = x + y$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+y}$$ A first characteristic equation from solving $\frac{dx}{2x}=\frac{dy}{3y}$ that you found correctly : $$\frac{x^3}{y^2}=c_1$$ A second characteristic equation from $$\frac{dx}{2x}=\frac{dy}{3y}=\frac{dz}{x+y}=\frac{3dx+2dy-6dz}{3(2x)+2(3y)-6(x+y)}=\frac{3dx+2dy-6dz}{0}$$ See explanation below. $$\implies\quad 3dx+2dy-6dz=0$$ $$z-\frac12 x-\frac13 y=c_2$$ General solution of the PDE on the inplicit form $c_2=f(c_1)$ $$z-\frac12 x-\frac13 y=f\left(\frac{x^3}{y^2}\right)$$ $$\boxed{z(x,y)=\frac12 x+\frac13 y+f\left(\frac{x^3}{y^2}\right)}$$
Explanation about combining fractions :
Use the well known basic property of the fractions :
$$\text{if}\quad \frac{A}{B}=\frac{C}{D} \quad \text{then}\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$ $c_1$ , $c_2$ are arbitrary constants (not both nul}.
This property is valid for more fractions :
$$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$ In above case $c_1=3\:;\:c_2=2 \:;\: c_3=-6 $ .
