Question

U(x+y) U_x + U (x-y) U_y = x² + y²

Answer 1

$u(x+y)\frac{\partial u}{\partial x}+u(x-y)\frac{\partial u}{\partial y}=x^2+y^2\\ v(u,x,y)=0\\ \frac{\partial v}{\partial u}\neq 0\\ u(x+y)\frac{\partial v}{\partial x}+u(x-y)\frac{\partial v}{\partial y}+(x^2+y^2)\frac{\partial v}{\partial u}=0\\ \frac{dx}{u(x+y)}=\frac{dy}{u(x-y)}=\frac{du}{x^2+y^2}\\ \frac{dx}{u(x+y)}=\frac{dy}{u(x-y)}\\ y^{\prime}=\frac{x-y}{x+y}\\ y^{\prime}=1-\frac{2y}{x}\cdot\frac{1}{1+\frac{y}{x}}\\ z(x)=\frac{y}{x};y=z\cdot x;y^{\prime}=z^{\prime}\cdot x+z\\ z^{\prime}\cdot x+z=1-2z\cdot\frac{1}{1+z}\\ z^{\prime}\cdot x+z+2z\cdot\frac{1}{1+z}-1=0\\ \frac{dz}{dx}\cdot x+z+2z\cdot\frac{1}{1+z}-1=0\\ xdz+zdx+2z\cdot\frac{1}{1+z}dx-dx=0\\ xdz+(z+2z\cdot\frac{1}{1+z}-1)dx=0\\ xdz+\frac{z^2+2z-1}{z+1}dx=0|:x\frac{z^2+2z-1}{z+1}\neq 0\\ \frac{(z+1)dz}{z^2+2z-1}+\frac{dx}{x}=0\\ \frac{1}{2}\cdot\frac{d(z^2+2z+1)}{z^2+2z-1}+\frac{dx}{x}=0\\ \frac{1}{2}\ln|z^2+2z-1|+\ln|x|=\ln C\\ \ln x\sqrt{z^2+2z-1}=\ln C\\ x\sqrt{z^2+2z-1}=C.\\ \frac{dx}{u(x+y)}=\frac{dy}{u(x-y)}\\\\ \frac{xdx-ydy}{ux(x+y)-uy(x-y)}=\frac{du}{x^2+y^2}\\ \frac{1}{2}\cdot\frac{d(x^2-y^2)}{u(x^2+y^2)}=\frac{du}{x^2+y^2}\\ \frac{1}{2}\cdot d(x^2-y^2)=udu\\ \frac{1}{2}\cdot (x^2-y^2)-\frac{u^2}{2}=C\\ x^2-y^2-u^2=C\\ f(x\sqrt{(y/x)^2+2(y/x)-1},\ x^2-y^2-u^2)=0\text{ --- general solution of our equation}.$