Question Number 208303 by Simurdiera last updated on 10/Jun/24
$${Resolver} \\ $$$$\frac{\partial^{\mathrm{2}} {u}}{\partial{y}^{\mathrm{2}} }\:−\:{x}^{\mathrm{2}} {u}\:=\:{xe}^{\mathrm{4}{y}} \\ $$
Answered by aleks041103 last updated on 12/Jun/24
$${u}={X}\left({x}\right){Y}\left({y}\right) \\ $$$$\Rightarrow{XY}\:''−{x}^{\mathrm{2}} {XY}={x}\:{e}^{\mathrm{4}{y}} \\ $$$$\frac{{Y}\:''}{{Y}}=\frac{{x}}{{X}}\:\frac{{e}^{\mathrm{4}{y}} }{{Y}}+{x}^{\mathrm{2}} \\ $$$${left}\:{side}\:{is}\:{function}\:{of}\:{y}\:{only} \\ $$$${right}\:{side}\:{has}\:{dependence}\:{on}\:{x} \\ $$$$\Rightarrow\frac{{Y}\:''}{{Y}}=\frac{{x}}{{X}}\:\frac{{e}^{\mathrm{4}{y}} }{{Y}}+{x}^{\mathrm{2}} ={c}={const}. \\ $$$${Y}=\frac{\mathrm{1}}{{a}}{e}^{\mathrm{4}{y}} \Rightarrow{c}=\mathrm{16} \\ $$$$\Rightarrow\mathrm{16}=\frac{{ax}}{{X}}+{x}^{\mathrm{2}} \Rightarrow{X}=\frac{{ax}}{\mathrm{16}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{u}={XY}=\frac{{xe}^{\mathrm{4}{y}} }{\mathrm{16}−{x}^{\mathrm{2}} } \\ $$$${the}\:{PDE}\:{is}\:{linear} \\ $$$$\Rightarrow{u}=\frac{{xe}^{\mathrm{4}{y}} }{\mathrm{16}−{x}^{\mathrm{2}} }+{w} \\ $$$${where} \\ $$$$\frac{\partial^{\mathrm{2}} {w}}{\partial{y}^{\mathrm{2}} }\:−\:{x}^{\mathrm{2}} {w}\:=\:\mathrm{0} \\ $$$$\Rightarrow{w}={f}\left({x}\right){e}^{{xy}} +{g}\left({x}\right){e}^{−{xy}} \\ $$$${For}\:{Cauchy}\:{problem}: \\ $$$${w}\left({x},\mathrm{0}\right)={w}_{\mathrm{0}} \left({x}\right)\:{and}\:\frac{\partial{w}}{\partial{y}}\left({x},\mathrm{0}\right)={v}_{\mathrm{0}} \left({x}\right) \\ $$$${we}\:{can}\:{solve}\:{for}\:{f}\:{and}\:{g}. \\ $$$${Therefore}\:{this}\:{is}\:{the}\:{general}\:{solution}. \\ $$$$\Rightarrow{u}\left({x},{y}\right)=\:\frac{{xe}^{\mathrm{4}{y}} }{\mathrm{16}−{x}^{\mathrm{2}} }\:+\:{f}\left({x}\right){e}^{{xy}} +{g}\left({x}\right){e}^{−{xy}} \\ $$