Answer to Question #140583 in Differential Equations for Nikhil Singh

Question

Solve
2x^2(∂^2/∂x^2) - 5xy[∂^2z/(∂x.∂y)] + 2y^2(∂^2z/∂y^2) + 2[x(∂z/∂x)+y(∂z/∂y) = 0

Answer 1

$2x^2\dfrac{\partial^2 z}{\partial x^2}-5xy\dfrac{\partial^2 z}{\partial x \partial y} + 2 y^2\dfrac{\partial^2 z}{\partial y^2} + 2[ x \dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}]=0\\$

Reduce the equation to the canonical form:

$A=2x^2\\ B=-2.5xy\\ C=2y^2\\ \Delta= B^2-AC=6.25x^2y^2-4x^2y^2=2.25 y^2$

So, we have hyperbolic type of equation.

Now let introduce a replacement.

$2x^2(dy)^2+5xy(dxdy)+2y^2(dx)^2=0|divide \space by \space (dx)^2\\ 2x^2(\dfrac{dy}{dx})^2 + 5xy(\dfrac{dy}{dx}) + 2y^2=0\\ y'_1=\dfrac{-5xy+\sqrt{D}}{4x^2}=\dfrac{-y}{2x}=> \eta=2x+y\\ y'_2=\dfrac{-5xy-\sqrt{D}}{4x^2}=\dfrac{-2y}{x}=>\psi=x+2y\\$

Now rewrite equation in new parameters:

$\dfrac{\partial z}{\partial x}=2\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \psi}\\ \dfrac{\partial z}{\partial y}=\dfrac{\partial z}{\partial \eta}+2\dfrac{\partial z}{\partial \psi}\\ \dfrac{\partial^2 z}{\partial x^2}=4\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+\dfrac{\partial^2 z}{\partial \psi^2}\\ \dfrac{\partial^2 z}{\partial y^2}=\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+4\dfrac{\partial^2 z}{\partial \psi^2}\\ \dfrac{\partial^2 z}{\partial x \partial y}=2\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+2\dfrac{\partial^2 z}{\partial \psi^2}\\ Our \space equation:\\ \dfrac{8\eta^2 -8\eta\psi +2\psi^2}{9}(4z_{\eta\eta}+4z_{\eta \psi}+z_{\psi \psi})-5\dfrac{5\psi\eta-2\psi^2-2\eta^2}{9}(2\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+2\dfrac{\partial^2 z}{\partial \psi^2})+\dfrac{8\psi^2 -8\eta\psi +2\eta^2}{9}(\dfrac{\partial^2 z}{\partial \eta^2}+4\dfrac{\partial^2 z}{\partial \psi \partial \eta}+4\dfrac{\partial^2 z}{\partial \psi^2})+2[\dfrac{2\eta-\psi}{3}(2\dfrac{\partial z}{\partial \eta}+\dfrac{\partial z}{\partial \psi})+\dfrac{2\psi-\eta}{3}(\dfrac{\partial z}{\partial \eta}+2\dfrac{\partial z}{\partial \psi})=0$