Answer to Question #150896 in Differential Equations for Ashweta Padhan

Question

Find the integral surface of the differential equation x(z+2)p+(xz+2yz+2y)q=z(z+1) passing through the curve x0= s^2, y0= 0 and z0 = 2s

Answer 1

$\frac{dx}{x(z+2)}=\frac{dy}{xz+2yz+2y}=\frac{dz}{z(z+1)}$

$\frac{dx}{x(z+2)}=\frac{dz}{z(z+1)}$

$\frac{dx}{x}=\frac{(z+2)dz}{z(z+1)}$

$\intop\frac{dx}{x}=\intop\frac{(z+2)dz}{z(z+1)}$

$\intop\frac{(z+2)dz}{z(z+1)}=\intop\frac{dz}{z+1}+2\intop\frac{dz}{z(z+1)}$

$\frac{1}{z(z+1)}=\frac{A}{z}+\frac{B}{z+1}$

$A(z+1)+Bz=1$

$A+B=0, A=1, B=-1$

$\frac{1}{z(z+1)}=\frac{1}{z}-\frac{1}{z+1}$

$lnx=ln(z+1)+2lnz-2ln(z+1)+lnc_1$

$lnx=ln(\frac{c_1z^2}{z+1})$

$\frac{x(z+1)}{z^2}=c_1$

$\frac{dy}{xz+2yz+2y}=\frac{dz}{z(z+1)}$

$\frac{(z+1)dy}{c_1z^3+2y(z+1)^2}=\frac{dz}{z(z+1)}$

$\frac{dy}{dz}-\frac{y}{z}=\frac{c_1z^2}{(z+1)^2}$

$y=uv, y'=u'v+uv'$

$u'v+u(v'-\frac{v}{z})=\frac{c_1z^2}{(z+1)^2}$

$v'-\frac{v}{z}=0\implies v=z$

$u'v=u'z=\frac{c_1z^2}{(z+1)^2}$

$\frac{du}{dz}=\frac{c_1z}{(z+1)^2}$

$u=c_1\intop\frac{zdz}{(z+1)^2}$

$\frac{z}{(z+1)^2}=\frac{A}{z+1}+\frac{B}{(z+1)^2}$

$A(z+1)+B=z$

$A=1, A+B=0, B=-1$

$\frac{z}{(z+1)^2}=\frac{1}{z+1}-\frac{1}{(z+1)^2}$

$u=c_1\intop(\frac{1}{z+1}-\frac{1}{(z+1)^2})dz=c_1(ln(z+1)+\frac{1}{z+1})+c_2$

$y=uv=c_1z(ln(z+1)+\frac{1}{z+1})+c_2z$

$y=\frac{x(z+1)}{z}(ln(z+1)+\frac{1}{z+1})+c_2z$

$\frac{y}{z}-\frac{x(z+1)}{z^2}(ln(z+1)+\frac{1}{z+1})=c_2$

For the curve $(s^2, 0, 2s)$ :

$c_1=\frac{s^2(2s+1)}{4s^2}=\frac{2s+1}{4}$

$c_2=-\frac{2s+1}{4}(ln(2s+1)+\frac{1}{2s+1})$

$c_2=-c_1(ln(4c_1)+\frac{1}{4c_1})=-c_1ln(4c_1)-\frac{1}{4}$

Substituting $c_1$ and $c_2$ we get the integral surface:

$\frac{y}{z}-\frac{x(z+1)}{z^2}(ln(z+1)+\frac{1}{z+1})=-\frac{x(z+1)}{z^2}ln(\frac{4x(z+1)}{z^2})-\frac{1}{4}$

$\frac{y}{z}-\frac{x}{z^2}=-\frac{x(z+1)}{z^2}ln(\frac{4x}{z^2})-\frac{1}{4}$