Answer to Question #216342 in Differential Equations for Rakesh

Question

(a) Find the integral surface of the following partial differential equation: (6) x(z+2a)p+ (xz+2yz + 2ay)q=z(z + a) Also find a particular solution passing through y = 0, 23+x(z + a)

Answer 1

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

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

$\int\frac{dx}{x}=\intop\frac{dz}{z}+\int\frac{adz}{z(z+a)}$

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

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

$A+B=0$

A=1

B=-1

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

$log(x)=log(z)-a log(z+a)+ log(c_1)=log(\frac{c_1-z^2}{z+a})$

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

$\frac{dy}{xz+2yz+2ay}=\frac{dz}{z(z+a)}$

$\frac{(z+a)dy}{c_1-z^3+2y(z+1)^2}=\frac{dz}{z(z+a)}$

$\frac{dy}{dz}-\frac{y}{z}=\frac{c_1-z^2}{(z+a)^2}$

$y=uv$

$y'=u'v+v'y$

$v'-v/z=0$

v=z

$\frac{dy}{dz}=\frac{c_1-z^2}{(z+a)^2}$

$c_1=log(z+a)+\frac{1}{z+a}+c_2-z$

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

For y=0.23+x(r+a)

$c_2=\frac{0.23+x(z+a)}{z}-\frac{x(z+a)}{z^2}-log(z+a)+\frac{1}{z+a}$

Substituting c₁ and c₂ we get the integral surface:

$\frac{2x(z+a)}{z^2}=\frac{2}{z+a}+\frac{0.23+x(z+a)}{z}-z$