Answer to Question #190300 in Differential Equations for Utkarsh

Question

Solve the full problem

X²u_xx +2xyu_xy+u_yy=u

Answer 1

The given equation is-

$x^2u_{xx} +2xyu_{xy}+u_{yy}=u$

Change of variable method-

From this eqn.$b^2-4ac=4x^2y^2-4x^2y^2=0$ so eqn is parabolic

General form-

$au_{xx}+2bu_{xy}+cu_{yy}+du=0$

Charcterstics equation is-

$\dfrac{dy}{dx}=\dfrac{b}{a}=\dfrac{2xy}{x^2}=\dfrac{2y}{x}$

Let $h=x ,t=\dfrac{y}{x}$

$\Rightarrow u_x=u_x(1)+u_t(\dfrac{-y}{x^2})=u_x-\dfrac{y}{x^2}u_t u_y=u_h(0)+u_t(\dfrac{1}{x}=\dfrac{1}{x}u_t$

$u_{xx}=u_{hh}-\dfrac{2y}{x^2}u_{ht}+\dfrac{y^2}{x^4}u_{tt}+\dfrac{2y}{x^3}u_t~~~~~~~-(1)\\[9pt]u_{yy}=\dfrac{1}{x^2}u_{tt} ~~~~~~-(2)\\[9pt]u_{xy}=\dfrac{1}{x}u_{xt}-\dfrac{y}{x^3}u_{tt}-\dfrac{1}{x^2}u_t~~~~~~~~-(3)$

Substituting eqn. (1), (2) and (3) in given Pde and we get-

$h^2u_{hh}-2yu_{ht}+t^2u_{tt}+2tu_t+t^2u_{tt}+2yu_{ht}-2t^2u_{tt}-2tu_{t}=u$

$h^2u_{hh}=u$

$u_{hh}=\dfrac{u}{h^2}$

or $\dfrac{d^2u}{dh^2}=\dfrac{u}{h^2}$

Integrating and we get-

$logu\dfrac{du}{dh}=\dfrac{-1}{h}+f(t)$

Again Integrating and we get-

$u_{h,t}=-lnh+hf(t)+g(t)$

On substituting the values of h and t we get-

$u(x,y)=2x^2+xf(\dfrac{y}{x})+g(\dfrac{y}{x})$