Question Number 163591 by Ar Brandon last updated on 08/Jan/22
$$\mathrm{R}\acute {\mathrm{e}soudre}\:\:\:\:\:\:\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }=\mathrm{10}{e}^{\mathrm{2}{x}+{y}} \\ $$
Commented by mkam last updated on 08/Jan/22
$$\boldsymbol{{U}}_{{p}} \:=\:\frac{\mathrm{1}}{\boldsymbol{{D}}^{\mathrm{2}} \:+\:\boldsymbol{{D}}^{'} \:^{\mathrm{2}} }\:\boldsymbol{{e}}^{\boldsymbol{{ax}}\:+\:\boldsymbol{{by}}} \:\:\::\:\boldsymbol{{D}}^{\mathrm{2}} \:+\:\boldsymbol{{D}}^{'} \:^{\mathrm{2}} \:\neq\:\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{y}}} \:\Rightarrow\:\boldsymbol{{a}}\:=\:\mathrm{2}\:,\:\boldsymbol{{b}}\:=\:\mathrm{1} \\ $$$$ \\ $$$$\boldsymbol{{D}}\left(\boldsymbol{{a}}\:,\:\boldsymbol{{b}}\:\right)\:=\:\boldsymbol{{D}}\:\left(\:\mathrm{2}\:,\:\mathrm{1}\:\right)\:=\:\mathrm{2}^{\mathrm{2}} \:+\:\mathrm{1}^{\mathrm{2}} \:=\:\mathrm{5}\:\neq\mathrm{0} \\ $$$$ \\ $$$$\therefore\boldsymbol{{U}}_{{p}} \:=\:\frac{\mathrm{10}}{\mathrm{5}}\:\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{y}}} \:=\:\mathrm{2}\:\boldsymbol{{e}}^{\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{y}}} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 08/Jan/22
Thank you ����
Commented by mkam last updated on 08/Jan/22
$$\boldsymbol{{you}}\:\boldsymbol{{are}}\:\boldsymbol{{welcome}}\:\boldsymbol{{sir}} \\ $$