Menu Close

dy-dt-3t-2-y-t-2-y-0-1-y-t-




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

Leave a Reply

Your email address will not be published. Required fields are marked *