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Question Number 108005 by Ar Brandon last updated on 13/Aug/20
1. x′′(t)+x(t)=tcos(2t)+(1+t^2 )sin(2t) 2. x′′(t)+x(t)=t^2 cos(2t)
$$\mathrm{1}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{tcos}\left(\mathrm{2t}\right)+\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{2t}\right) \\ $$$$\mathrm{2}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right) \\ $$
Answered by mathmax by abdo last updated on 13/Aug/20
2) y^(′′) +y =t^2 cos(2t) h→r^2 +1 =0 ⇒r =+^− i ⇒y =acost +bsint =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((cost sint)),((−sint cost)))=1 ≠0 W_1 = determinant (((o sint)),((t^2 cos(2t) cost)))=−t^2 cos(2t)sint W_2 = determinant (((cost 0)),((−sint t^2 cos(2t))))=t^2 cost .cos(2t) v_1 =∫ (w_1 /W)dt =−∫ t^2 cos(2t)sint dt we have cos(2t)sint =cos(2t)cos((π/2)−t) =(1/2){cos(t+(π/2))+cos(3t−(π/2))} =(1/2){sin(3t)−sint} ⇒v_1 =−(1/2)(∫ t^2 sin(3t)dt−∫ t^2 sint dt) =(1/2)∫ t^2 sint dt −(1/2)∫t^2 sin(3t)dt but ∫ t^2 sint dt =−t^2 cost +∫ (2t) cost dt =−t^2 cost +2{ tsint −∫ sint dt} =−t^2 cost +2t sint +2cost ∫ t^2 sin(3t)dt =_(3t =x) ∫ (x^2 /9)sinx (dx/3) =(1/(27)) ∫x^2 sinx dx =(1/(27)){−x^2 cosx +2x sinx +2cosx} =(1/(27)){−(3t)^2 cos(3t)+6t sin(3t)+2cos(3t)} ⇒ v_1 =(1/2){−t^2 cost +2tsint +2cost} −(1/(54)){−9t^2 cos(3t)+6t sin(3t)+2cos(3t)} v_2 =∫ (w_2 /W)dt =∫ t^2 cost cos(2t)dt =(1/2)∫ t^2 {cos(3t)+cost} dt =(1/2)∫ t^2 cos(3t)dt +(1/2)∫ t^2 cost dt ∫ t^2 cost dt =t^2 sint −∫ 2t sint dt =t^2 sint −2{−tcost +∫ cost dt} =t^2 sint +2tcost −2cost ∫ t^2 cos(3t)dt =_(3t =u) ∫ (u^2 /9)cos(u)(du/3) =(1/(27))∫ u^2 cosu du =(1/(27)){u^2 sinu +2u cosu−2cosu}=(1/(27)){9t^2 sin(3t)+6t cos(3t)−2cos(3t)} ⇒y_p =u_1 v_1 +u_2 v_2 and general solution is y =y_h +y_p
$$\left.\mathrm{2}\right)\:\mathrm{y}^{''} \:+\mathrm{y}\:=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right) \\ $$$$\mathrm{h}\rightarrow\mathrm{r}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\mathrm{r}\:=\overset{−} {+}\mathrm{i}\:\Rightarrow\mathrm{y}\:=\mathrm{acost}\:+\mathrm{bsint}\:=\mathrm{au}_{\mathrm{1}} \:+\mathrm{bu}_{\mathrm{2}} \\ $$$$\mathrm{W}\left(\mathrm{u}_{\mathrm{1}} ,\mathrm{u}_{\mathrm{2}} \right)\:=\begin{vmatrix}{\mathrm{cost}\:\:\:\:\:\:\:\:\:\:\mathrm{sint}}\\{−\mathrm{sint}\:\:\:\:\:\:\:\mathrm{cost}}\end{vmatrix}=\mathrm{1}\:\neq\mathrm{0} \\ $$$$\mathrm{W}_{\mathrm{1}} =\begin{vmatrix}{\mathrm{o}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{sint}}\\{\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right)\:\:\:\:\:\mathrm{cost}}\end{vmatrix}=−\mathrm{t}^{\mathrm{2}} \:\mathrm{cos}\left(\mathrm{2t}\right)\mathrm{sint} \\ $$$$\mathrm{W}_{\mathrm{2}} =\begin{vmatrix}{\mathrm{cost}\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}}\\{−\mathrm{sint}\:\:\:\:\:\:\:\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right)}\end{vmatrix}=\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:.\mathrm{cos}\left(\mathrm{2t}\right) \\ $$$$\mathrm{v}_{\mathrm{1}} =\int\:\frac{\mathrm{w}_{\mathrm{1}} }{\mathrm{W}}\mathrm{dt}\:=−\int\:\mathrm{t}^{\mathrm{2}} \:\mathrm{cos}\left(\mathrm{2t}\right)\mathrm{sint}\:\mathrm{dt}\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{cos}\left(\mathrm{2t}\right)\mathrm{sint}\:=\mathrm{cos}\left(\mathrm{2t}\right)\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{cos}\left(\mathrm{t}+\frac{\pi}{\mathrm{2}}\right)+\mathrm{cos}\left(\mathrm{3t}−\frac{\pi}{\mathrm{2}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{sin}\left(\mathrm{3t}\right)−\mathrm{sint}\right\}\:\Rightarrow\mathrm{v}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\left(\int\:\mathrm{t}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{3t}\right)\mathrm{dt}−\int\:\mathrm{t}^{\mathrm{2}} \mathrm{sint}\:\mathrm{dt}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{t}^{\mathrm{2}} \mathrm{sint}\:\mathrm{dt}\:−\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{t}^{\mathrm{2}} \:\mathrm{sin}\left(\mathrm{3t}\right)\mathrm{dt}\:\:\mathrm{but} \\ $$$$\int\:\mathrm{t}^{\mathrm{2}} \:\mathrm{sint}\:\mathrm{dt}\:=−\mathrm{t}^{\mathrm{2}} \:\mathrm{cost}\:+\int\:\left(\mathrm{2t}\right)\:\mathrm{cost}\:\mathrm{dt} \\ $$$$=−\mathrm{t}^{\mathrm{2}} \:\mathrm{cost}\:+\mathrm{2}\left\{\:\mathrm{tsint}\:−\int\:\mathrm{sint}\:\mathrm{dt}\right\}\:=−\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:+\mathrm{2t}\:\mathrm{sint}\:\:+\mathrm{2cost} \\ $$$$\int\:\mathrm{t}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{3t}\right)\mathrm{dt}\:=_{\mathrm{3t}\:=\mathrm{x}} \:\:\:\int\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{9}}\mathrm{sinx}\:\frac{\mathrm{dx}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{27}}\:\int\mathrm{x}^{\mathrm{2}} \mathrm{sinx}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\left\{−\mathrm{x}^{\mathrm{2}} \mathrm{cosx}\:+\mathrm{2x}\:\mathrm{sinx}\:+\mathrm{2cosx}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\left\{−\left(\mathrm{3t}\right)^{\mathrm{2}} \mathrm{cos}\left(\mathrm{3t}\right)+\mathrm{6t}\:\mathrm{sin}\left(\mathrm{3t}\right)+\mathrm{2cos}\left(\mathrm{3t}\right)\right\}\:\Rightarrow \\ $$$$\mathrm{v}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left\{−\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:+\mathrm{2tsint}\:+\mathrm{2cost}\right\}\:−\frac{\mathrm{1}}{\mathrm{54}}\left\{−\mathrm{9t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{3t}\right)+\mathrm{6t}\:\mathrm{sin}\left(\mathrm{3t}\right)+\mathrm{2cos}\left(\mathrm{3t}\right)\right\} \\ $$$$\mathrm{v}_{\mathrm{2}} =\int\:\frac{\mathrm{w}_{\mathrm{2}} }{\mathrm{W}}\mathrm{dt}\:=\int\:\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:\mathrm{cos}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{t}^{\mathrm{2}} \left\{\mathrm{cos}\left(\mathrm{3t}\right)+\mathrm{cost}\right\}\:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{3t}\right)\mathrm{dt}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\:\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:\mathrm{dt} \\ $$$$\int\:\mathrm{t}^{\mathrm{2}} \mathrm{cost}\:\mathrm{dt}\:=\mathrm{t}^{\mathrm{2}} \mathrm{sint}\:−\int\:\mathrm{2t}\:\mathrm{sint}\:\mathrm{dt}\:\:=\mathrm{t}^{\mathrm{2}} \mathrm{sint}\:−\mathrm{2}\left\{−\mathrm{tcost}\:+\int\:\mathrm{cost}\:\mathrm{dt}\right\} \\ $$$$=\mathrm{t}^{\mathrm{2}} \mathrm{sint}\:+\mathrm{2tcost}\:−\mathrm{2cost} \\ $$$$\int\:\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{3t}\right)\mathrm{dt}\:=_{\mathrm{3t}\:=\mathrm{u}} \:\:\:\int\:\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{9}}\mathrm{cos}\left(\mathrm{u}\right)\frac{\mathrm{du}}{\mathrm{3}}\:=\frac{\mathrm{1}}{\mathrm{27}}\int\:\mathrm{u}^{\mathrm{2}} \mathrm{cosu}\:\mathrm{du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\left\{\mathrm{u}^{\mathrm{2}} \mathrm{sinu}\:+\mathrm{2u}\:\mathrm{cosu}−\mathrm{2cosu}\right\}=\frac{\mathrm{1}}{\mathrm{27}}\left\{\mathrm{9t}^{\mathrm{2}} \mathrm{sin}\left(\mathrm{3t}\right)+\mathrm{6t}\:\mathrm{cos}\left(\mathrm{3t}\right)−\mathrm{2cos}\left(\mathrm{3t}\right)\right\} \\ $$$$\:\Rightarrow\mathrm{y}_{\mathrm{p}} =\mathrm{u}_{\mathrm{1}} \mathrm{v}_{\mathrm{1}} \:+\mathrm{u}_{\mathrm{2}} \mathrm{v}_{\mathrm{2}} \:\mathrm{and}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{y}\:=\mathrm{y}_{\mathrm{h}} \:+\mathrm{y}_{\mathrm{p}} \\ $$$$ \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 14/Aug/20
Thanks
Commented by abdomathmax last updated on 14/Aug/20
you are welcome
$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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