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Question Number 219488 by OmoloyeMichael last updated on 26/Apr/25
solve the initial value problem y′−2e^(−t^2 ) +2ty=0 y(0)=1
$$\boldsymbol{{solve}}\:\boldsymbol{{the}}\:\boldsymbol{{initial}}\:\boldsymbol{{value}}\:\boldsymbol{{problem}}\: \\ $$$$\boldsymbol{{y}}'−\mathrm{2}\boldsymbol{{e}}^{−\boldsymbol{{t}}^{\mathrm{2}} } +\mathrm{2}\boldsymbol{{ty}}=\mathrm{0}\:\:\boldsymbol{{y}}\left(\mathrm{0}\right)=\mathrm{1} \\ $$
Answered by SdC355 last updated on 26/Apr/25
(dy/dt)+2ty(t)=2e^(−t^2 ) Let μ(t)=e^(∫ 2t dt) integrating Factor e^t^2 (dy/dt)+2te^t^2 y(t)=2 Multiple integrating factor each side ((d )/dt)(e^t^2 y(t))=2 (∵ f′g+fg′=(fg)′) ∫ 2dt=e^t^2 y(t)+C 2t+C_1 =e^t^2 y(t)+C 2t+C′=e^t^2 y(t) y(t)=2te^(−t^2 ) +C′e^(−t^2 ) , y(0)=1 y(t)=(2t+1)e^(−t^2 )
$$\frac{\mathrm{d}{y}}{\mathrm{d}{t}}+\mathrm{2}{ty}\left({t}\right)=\mathrm{2}{e}^{−{t}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:\mu\left({t}\right)={e}^{\int\:\mathrm{2}{t}\:\mathrm{d}{t}} \:\:\mathrm{integrating}\:\mathrm{Factor} \\ $$$${e}^{{t}^{\mathrm{2}} } \frac{\mathrm{d}{y}}{\mathrm{d}{t}}+\mathrm{2}{te}^{{t}^{\mathrm{2}} } {y}\left({t}\right)=\mathrm{2}\:\:\mathrm{Multiple}\:\mathrm{integrating}\:\mathrm{factor}\:\mathrm{each}\:\mathrm{side} \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\left({e}^{{t}^{\mathrm{2}} } {y}\left({t}\right)\right)=\mathrm{2}\:\:\:\left(\because\:{f}'\mathrm{g}+{f}\mathrm{g}'=\left({f}\mathrm{g}\right)'\right) \\ $$$$\int\:\mathrm{2d}{t}={e}^{{t}^{\mathrm{2}} } {y}\left({t}\right)+{C} \\ $$$$\mathrm{2}{t}+{C}_{\mathrm{1}} ={e}^{{t}^{\mathrm{2}} } {y}\left({t}\right)+{C} \\ $$$$\mathrm{2}{t}+{C}'={e}^{{t}^{\mathrm{2}} } {y}\left({t}\right) \\ $$$${y}\left({t}\right)=\mathrm{2}{te}^{−{t}^{\mathrm{2}} } +{C}'{e}^{−{t}^{\mathrm{2}} } \:\:,\:{y}\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${y}\left({t}\right)=\left(\mathrm{2}{t}+\mathrm{1}\right){e}^{−{t}^{\mathrm{2}} } \\ $$
Commented by OmoloyeMichael last updated on 26/Apr/25
thanks sir
$$\boldsymbol{{thanks}}\:\boldsymbol{{sir}} \\ $$

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