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Question Number 199891 by Mathspace last updated on 10/Nov/23
$${solve}\:{by}\:{laplce}\:{transform} \\ $$$${y}^{''} −{y}^{'} +{y}\:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$
Commented by Mathspace last updated on 10/Nov/23
$${with}\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:{and}\:{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1} \\ $$