Find the solution to the following differential equation. Take the constants of integration to be zeros
xy′+(1+2x2)y=x3e−x2\displaystyle {x}{y}'+{\left({1}+{2}{x}^{{2}}\right)}{y}={x}^{{3}}{e}^{{-{x}^{{2}}}}xy′+(1+2x2)y=x3e−x2
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