Question

solve (x^3+3xy^2)p+(y^3+3x^2y)q=(x+y)^2z

Answer 1

Given equation is-

$(x^3+3xy^2)p+(y^3+3x^2y)q=(x+y)^2z$

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

$\dfrac{dy}{dx}=\dfrac{y^3+3x^2y}{x^3+3xy^2}$

This is a homogeneous differential equation.

Let $\dfrac{y}{x}=v, y=xv, \dfrac{dy}{dx}=v+x(\dfrac{dv}{dx})$

Then:

$v+x\dfrac{dv}{dx}=\dfrac{(xv)^3+3x^3v}{x^3+3x^3v^2}=\dfrac{v^3+3v}{1+3v^2}$

$x\dfrac{dv}{dx}=\dfrac{v^3+3v}{1+3v^2}-v=\dfrac{4v^3+2v}{1+3v^2}$

$\dfrac{1}{2}\intop\dfrac{1+3v^2}{v(2v^2+1)}dv=\intop \dfrac{dx}{x}$

$\dfrac{1+3v^2}{v(2v^2+1)}=\dfrac{A}{v}+\dfrac{Bv+C}{2v^2+1}$

$A(2v^2+1)+v(Bv+C)=1+3v^2$

$2A+B=3$

$A=1$

$C=0$

$B=3-2A=1$

$\dfrac{1-3v^2}{v(2v^2+1)}=\dfrac{1}{v}+\dfrac{v}{2v^2+1}$

$\intop\dfrac{v}{2v^2+1}dv=\dfrac{1}{2}\intop\dfrac{d(v^2)}{2v^2+1}dv=\dfrac{ln(2v^2+1)}{4}$

$\dfrac{lnv}{2}-\dfrac{5ln(2v^2+1)}{8}=lnx+lnc_1$

$ln(\dfrac{v^{1/2}}{(2v^2+1)^{5/8}})=ln(c_1x)$

$\dfrac{(y/x)^{1/2}}{x(2(y/x)^2+1)^{5/8}}=c_1$

$\dfrac{dx/x+dy/y+dz/z}{6x^2}=-\dfrac{3(dx/x-dy/y+dz/z)}{6y^2}=\dfrac{dz}{z(x+y)^2}$

$\dfrac{dx/x+dy/y+dz/z-3(dx/x-dy/y+dz/z)}{6x^2+6y^2}=\dfrac{3dz/z}{6(x+y)^2}$

$\dfrac{dx}{x}+\dfrac{dy}{y}+\dfrac{dz}{z}-3(\dfrac{dx}{x}-\dfrac{dy}{y}+\dfrac{dz}{z})=3\dfrac{dz}{z}$

$4\dfrac{dy}{y}-2\dfrac{dx}{x}=5\dfrac{dz}{z}$

$4lny-2lnx=5lnz+lnc_2$

$ln(\dfrac{y^4}{x^2})=ln(c_2z^5)$

$\dfrac{y^4}{x^2z^5}=c_2$

The general integral:

$\phi(\dfrac{(y/x)^{1/2}}{x(2(y/x)^2+1)^{5/8}}, \dfrac{y^4}{x^2z^5})=0$