Answer to Question #150978 in Differential Equations for Sachin

Question

Simultanius ordinary diff. Eqn dx/x^2-y^2-z^2=dy/2xy=dz/2zx

Answer 1

$\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2zx}$

$\frac{dy}{y}=\frac{dz}{z}$

$lny=ln(c_1z)$

$\frac{y}{z}=c_1$

$\frac{dx}{x^2-(c_1z)^2-z^2}=\frac{dz}{2zx}$

$\frac{2zx}{x^2-(c_1z)^2-z^2}=\frac{dz}{dx}$

This is a homogeneous DE.

$z=tx, z'=t'x+t$

$t'x+t=\frac{2tx^2}{x^2-t^2x^2(c_1+1)}=\frac{2t}{1-t^2(c_1+1)}$

$t'x=\frac{2t}{1-t^2(c_1+1)}-t=\frac{t+t^3(c_1+1)}{1-t^2(c_1+1)}$

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

$\frac{1-t^2(c_1+1)}{t+t^3(c_1+1)}=\frac{1-t^2(c_1+1)}{t(1+t(c_1+1)^{1/3})(1-t(c_1+1)^{1/3}+t^2(c_1+1)^{2/3})}$

$\frac{1-t^2(c_1+1)}{t(1+t(c_1+1)^{1/3})(1-t(c_1+1)^{1/3}+t^2(c_1+1)^{2/3})}=\frac{A}{t}+\frac{B}{1+t(c_1+1)^{1/3}}+\frac{Ct+D}{1-t(c_1+1)^{1/3}+t^2(c_1+1)^{2/3}}$

$A(1+t^3(c_1+1))+Bt(1-t(c_1+1)^{1/3}+t^2(c_1+1)^{2/3})+(Ct+D)t(1+t(c_1++1)^{1/3})=1-t^2(c_1+1)$

$A+At^3(c_1+1)+Bt-Bt^2(c_1+1)^{1/3}+Bt^3(c_1+1)^{2/3}+Ct^2+Ct^3(c_1+1)^{1/3}+$

$+Dt+Dt^2(c_1+1)^{1/3}=1-t^2(c_1+1)$

$A=1$

$B+D=0$

Let $(c_1+1=k^3)$ , then:

$-Bk+C+Dk=-k^3$

$C-2Bk=-k^3$

$Ak^3+Bk^2+Ck=0$

$C=-k^2-Bk$

$-k^2-Bk-2Bk=-k^3$

$B=\frac{k^2-k}{3}, D=\frac{k-k^2}{3}$

$C=\frac{-2k^2-k^3}{3}$

$\intop\frac{1-t^2(c_1+1)}{t+t^3(c_1+1)}dt=\intop(\frac{1}{t}+\frac{k^2-k}{3(1+kt)}+\frac{k-k^3-t(2k^2+k^3)}{3(1-kt+k^2t^2)})dt$

$\intop\frac{k-k^3-t(2k^2+k^3)}{3(1-kt+k^2t^2)}dt=\frac{k}{3}\intop\frac{1-k^2-kt(k+2)}{1-kt+k^2t^2}dt=$

$=\frac{k^2-1}{3\sqrt{3}}arctan{\frac{2kt}{\sqrt{3}}}-\frac{1}{6\sqrt{3}}ln(t^2-t/k+1/k^2)+\frac{1}{3k\sqrt{3}}arctan{\frac{2kt}{\sqrt{3}}}$

$lnt+\frac{k-1}{3}ln(1+kt)+\frac{k^2-1}{3\sqrt{3}}arctan{\frac{2kt}{\sqrt{3}}}-\frac{1}{6\sqrt{3}}ln(t^2-t/k+1/k^2)+\frac{1}{3k\sqrt{3}}arctan{\frac{2kt}{\sqrt{3}}}=$

$=lnx+c_2$

$ln(z/x)+\frac{k-1}{3}ln(1+kz/x)+\frac{k^2-1}{3\sqrt{3}}arctan{\frac{2kz}{x\sqrt{3}}}-\frac{1}{6\sqrt{3}}ln((z/x)^2-z/(kx)+1/k^2)++\frac{1}{3k\sqrt{3}}arctan{\frac{2kz}{x\sqrt{3}}}-lnx=c_2$