Question Number 197094 by MrGHK last updated on 07/Sep/23
Answered by sniper237 last updated on 07/Sep/23
$$\overset{{schwarz}\:{theo}} {\Rightarrow}\:\partial_{{x}} \left(\:{t}\partial_{{t}} {u}+\mathrm{2}{u}\right)={x}^{\mathrm{2}} \\ $$$$\overset{\int{dx}} {\Rightarrow}{t}\partial_{{t}} {u}+\mathrm{2}{u}\:=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{f}\left({t}\right) \\ $$$$\overset{{homogen}\:{slt\%}} {\Rightarrow}\:{u}_{{H}} =\frac{{C}}{{t}^{\mathrm{2}} } \\ $$$$\overset{{cte}\:{variat\%}} {\Rightarrow}\:\frac{{C}'\left({t}\right)}{{t}}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}+{f}\left({t}\right) \\ $$$$\:\:\Rightarrow\:{C}\left({t}\right)=\:\frac{{t}^{\mathrm{2}} {x}^{\mathrm{3}} }{\mathrm{6}}+\int{tf}\left({t}\right){dt} \\ $$$$\overset{{gen}\:{slt\%}} {\Rightarrow}\:\:{u}\left({x},{t}\right)=\frac{{x}^{\mathrm{3}} }{\mathrm{6}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int{tf}\left({t}\right){dt} \\ $$
Commented by MrGHK last updated on 07/Sep/23
$${sir}\:{can}\:{you}\:{explain}\:{how}\:{you}\:{get}\:{U}_{{H}} \:{sir} \\ $$
Commented by sniper237 last updated on 08/Sep/23
$${ty}'+\mathrm{2}{y}=\mathrm{0}\Rightarrow\frac{{y}'}{{y}}=−\frac{\mathrm{2}}{{t}}\Rightarrow\:{ln}\mid{y}\mid=−\mathrm{2}{ln}\mid{t}\mid+{k} \\ $$