Question Number 11887 by Peter last updated on 04/Apr/17
$$\frac{{dy}}{{dt}}\:+\mathrm{3}{t}^{\mathrm{2}} {y}\:=\:{t}^{\mathrm{2}\:} \:\:\:\:,\:{y}\left(\mathrm{0}\right)\:=\:\mathrm{1} \\ $$$${y}\left({t}\right)\:=\:? \\ $$
Answered by ajfour last updated on 04/Apr/17
$${when}\:\:\frac{{dy}}{{dt}}\:+{P}\:{y}\:={Q} \\ $$$${solution}\:{is}\::\: \\ $$$$\:\:{y}\left({e}^{\int{Pdt}} \right)\:=\:\int\left({Qe}^{\int{Pdt}} \right){dt} \\ $$$${here}\:\:{P}\:=\:\mathrm{3}{t}^{\mathrm{2}} \:\:\Rightarrow\:\:{e}^{\int{Pdt}} =\:{e}^{{t}^{\mathrm{3}} } \\ $$$$\:\:\:{y}\left({e}^{{t}^{\mathrm{3}} } \right)=\:\frac{\mathrm{1}}{\mathrm{3}}\int{e}^{{t}^{\mathrm{3}} } {d}\left({t}^{\mathrm{3}} \right) \\ $$$$\:\:\:{y}\left({e}^{{t}^{\mathrm{3}} } \right)\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left({e}^{{t}^{\mathrm{3}} } \right)^{\mathrm{2}} +{C} \\ $$$${as}\:{y}\left(\mathrm{0}\right)=\mathrm{1},\:\: \\ $$$$\:\:\:\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{6}}+{C}\:\:\:\Rightarrow\:\:\:{C}=\frac{\mathrm{5}}{\mathrm{6}}\: \\ $$$$\:\:{y}\:=\frac{\mathrm{5}+{e}^{\mathrm{2}{t}^{\mathrm{3}} } }{\mathrm{6}{e}^{{t}^{\mathrm{3}} } }\:. \\ $$