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Question Number 163099 by HongKing last updated on 03/Jan/22

1. u_(x x) = x + y 2. u_(x y) = x - y

$$\mathrm{1}.\:\mathrm{u}_{\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{x}}} \:=\:\boldsymbol{\mathrm{x}}\:+\:\boldsymbol{\mathrm{y}} \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{u}}_{\boldsymbol{\mathrm{x}}\:\boldsymbol{\mathrm{y}}} \:=\:\boldsymbol{\mathrm{x}}\:-\:\boldsymbol{\mathrm{y}} \\ $$

Answered by mr W last updated on 04/Jan/22

1) (∂^2 u/∂x^2 )=x+y (∂u/∂x)=(x^2 /2)+yx+f(y) ⇒u=(x^3 /6)+yx^2 +f(y)x+g(y) 2) (∂^2 u/(∂x∂y))=x−y (∂u/∂x)=xy−(y^2 /2)+f(x) u=((yx^2 )/2)−((y^2 x)/2)+∫f(x)dx+g(y)

$$\left.\mathrm{1}\right) \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }={x}+{y} \\ $$$$\frac{\partial{u}}{\partial{x}}=\frac{{x}^{\mathrm{2}} }{\mathrm{2}}+{yx}+{f}\left({y}\right) \\ $$$$\Rightarrow{u}=\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+{yx}^{\mathrm{2}} +{f}\left({y}\right){x}+{g}\left({y}\right) \\ $$$$\left.\mathrm{2}\right) \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{x}\partial{y}}={x}−{y} \\ $$$$\frac{\partial{u}}{\partial{x}}={xy}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{f}\left({x}\right) \\ $$$${u}=\frac{{yx}^{\mathrm{2}} }{\mathrm{2}}−\frac{{y}^{\mathrm{2}} {x}}{\mathrm{2}}+\int{f}\left({x}\right){dx}+{g}\left({y}\right) \\ $$

Commented by HongKing last updated on 05/Jan/22

thank you so much my dear Sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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