Question

Find the integral surface passing through the data curve x=1, z=y²+y of the equation x³ z_x+y(3x²+y)z_y =z(2x²+y)

Answer 1

we have to find the integral surface of the given equation:-

$x^3z_x+y(3x^2+y)z_y=z(2x^2+y)$

It is Lagrange's partial differential equation of type:

$Pp+Qq=R$

So, the

Lagrange's auxiliary equation will be

$\frac{dx}{x^3}=\frac{dy}{y(3x^2+y)}=\frac{dz}{z(2x^2+y)}$

At first we take

$\frac{dx}{x^3}=\frac{dy}{y(3x^2+y)}\\\implies\frac{dy}{dx}=\frac{y(3x^2+y)}{x^3}\\\implies\frac{1}{y^2}\frac{dy}{dx}=\frac{\frac{3x^2}{y}+1}{x^3}\\\implies-\frac{dt}{dx}=\frac{3t}{x}+\frac{1}{x^3}\\\implies\frac{dt}{dx}+\frac{3t}{x}=-\frac{1}{x^3}\\\\\\\\\implies tx^3=\intop(-\frac{1}{x^3})(x^3)dx\\\implies tx^3=\intop(-1)dx\\\implies tx^3=-x+c_1\\\implies\frac{x^3}{y}=-x+c_1.............(1)$

[ let $\frac{1}{y}=t ;$ & the integrating factor $e^(\intop\frac{3dx}{x})\\$

$\implies e^(\ln(x^3))\\\implies x^3$ ]

now we will take

$\frac{dx}{x^3}=\frac{dy}{y(3x^2+y)}=\frac{dz}{z(2x^2+y)}\\=\frac {zdy-ydz}{yz(3x^2+y)-yz(2x^2+y)}=\frac{zdy-ydz}{x^2yz}\\\implies\frac{dx}{x^3}=\frac{zdy-ydz}{x^2yz}\\\implies\frac{dx}{x}=\frac{zdy-ydz}{yz}\\\implies\frac{dx}{x}-\frac{dy}{y}+\frac{dz}{z}=0\\\implies\intop\frac{dx}{x}-\intop\frac{dy}{y}+\intop\frac{dz}{z}=0\\\implies\ln\frac{xz}{y}=\ln(c_2)\\\implies\frac{xz}{y}=c_2............(2)$

x=1& z=$y^2+y$ values are given

putting the values in the equations (1)&(2) we get

$c_1=\frac{1}{y}+1........(3)\\c_2=(y+1).......(4)$

putting the value of y of equation(4) in equation(3) we get

$(c_2-1)(c_1-1)=1\\\implies(\frac{xz}{y}-1)(\frac{x^3}{y}+x-1)=1......(5)$

hence the equation (5) is the required integral surface.