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Question Number 197094 by MrGHK last updated on 07/Sep/23

Answered by sniper237 last updated on 07/Sep/23

$$\stackrel{schwarztheo}{\Rightarrow }{\partial }_x(t{\partial }_{t}u+2u)=x^2$$$$\stackrel{\int dx}{\Rightarrow }t{\partial }_{t}u+2u=\frac{x^3}{3}+f\left(t\right)$$$$\stackrel{homogenslt\mathrm{\%}}{\Rightarrow }{u}_H=\frac{C}{t^2}$$$$\stackrel{ctevariat\mathrm{\%}}{\Rightarrow }\frac{C^{\prime }\left(t\right)}{t}=\frac{x^3}{3}+f\left(t\right)$$$$\Rightarrow C\left(t\right)=\frac{t^2{x}^3}{6}+\int tf\left(t\right)dt$$$$\stackrel{genslt\mathrm{\%}}{\Rightarrow }u(x,t)=\frac{x^3}{6}+\frac{1}{t^2}\int tf\left(t\right)dt$$

Commented by MrGHK last updated on 07/Sep/23

$$sircanyouexplainhowyouget{U}_{H}sir$$

Commented by sniper237 last updated on 08/Sep/23

$${ty}^{\prime }+2y=0\Rightarrow \frac{y^{\prime }}{y}=-\frac{2}{t}\Rightarrow ln\mid y\mid =-2ln\mid t\mid +k$$

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