1. x′′(t)+x(t)=tcos(2t)+(1+t^2 )sin(2t) 2. x′′(t)+x(t)=t^2 cos(2t)


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Question Number 108005 by Ar Brandon last updated on 13/Aug/20

$1 . x^{″} (t) + x (t) = tcos (2 t) + (1 + t^{2}) \sin (2 t)$ $2 . x^{″} (t) + x (t) = t^{2} \cos (2 t)$
	Answered by mathmax by abdo last updated on 13/Aug/20
	$2) y^(′′) +y =t^2 cos(2t) h→r^2 +1 =0 ⇒r =+^− i ⇒y =acost +bsint =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((cost sint)),((−sint cost)))=1 ≠0 W_1 = determinant (((o sint)),((t^2 cos(2t) cost)))=−t^2 cos(2t)sint W_2 = determinant (((cost 0)),((−sint t^2 cos(2t))))=t^2 cost .cos(2t) v_1 =∫ (w_1 /W)dt =−∫ t^2 cos(2t)sint dt we have cos(2t)sint =cos(2t)cos((π/2)−t) =(1/2){cos(t+(π/2))+cos(3t−(π/2))} =(1/2){sin(3t)−sint} ⇒v_1 =−(1/2)(∫ t^2 sin(3t)dt−∫ t^2 sint dt) =(1/2)∫ t^2 sint dt −(1/2)∫t^2 sin(3t)dt but ∫ t^2 sint dt =−t^2 cost +∫ (2t) cost dt =−t^2 cost +2{ tsint −∫ sint dt} =−t^2 cost +2t sint +2cost ∫ t^2 sin(3t)dt =_(3t =x) ∫ (x^2 /9)sinx (dx/3) =(1/(27)) ∫x^2 sinx dx =(1/(27)){−x^2 cosx +2x sinx +2cosx} =(1/(27)){−(3t)^2 cos(3t)+6t sin(3t)+2cos(3t)} ⇒ v_1 =(1/2){−t^2 cost +2tsint +2cost} −(1/(54)){−9t^2 cos(3t)+6t sin(3t)+2cos(3t)} v_2 =∫ (w_2 /W)dt =∫ t^2 cost cos(2t)dt =(1/2)∫ t^2 {cos(3t)+cost} dt =(1/2)∫ t^2 cos(3t)dt +(1/2)∫ t^2 cost dt ∫ t^2 cost dt =t^2 sint −∫ 2t sint dt =t^2 sint −2{−tcost +∫ cost dt} =t^2 sint +2tcost −2cost ∫ t^2 cos(3t)dt =_(3t =u) ∫ (u^2 /9)cos(u)(du/3) =(1/(27))∫ u^2 cosu du =(1/(27)){u^2 sinu +2u cosu−2cosu}=(1/(27)){9t^2 sin(3t)+6t cos(3t)−2cos(3t)} ⇒y_p =u_1 v_1 +u_2 v_2 and general solution is y =y_h +y_p$
	$2) y^{^{″}} + y = t^{2} \cos (2 t)$ $h \to r^{2} + 1 = 0 \Rightarrow r = \bar{+} i \Rightarrow y = acost + bsint = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} cost sint \\ - sint cost \end{matrix} \| = 1 \neq 0$ $W_{1} = \| \begin{matrix} o sint \\ t^{2} \cos (2 t) cost \end{matrix} \| = - t^{2} \cos (2 t) sint$ $W_{2} = \| \begin{matrix} cost 0 \\ - sint t^{2} \cos (2 t) \end{matrix} \| = t^{2} cost . \cos (2 t)$ $v_{1} = \int \frac{w_{1}}{W} dt = - \int t^{2} \cos (2 t) sint dt we have$ $\cos (2 t) sint = \cos (2 t) \cos (\frac{π}{2} - t) = \frac{1}{2} {\cos (t + \frac{π}{2}) + \cos (3 t - \frac{π}{2})}$ $= \frac{1}{2} {\sin (3 t) - sint} \Rightarrow v_{1} = - \frac{1}{2} (\int t^{2} \sin (3 t) dt - \int t^{2} sint dt)$ $= \frac{1}{2} \int t^{2} sint dt - \frac{1}{2} \int t^{2} \sin (3 t) dt but$ $\int t^{2} sint dt = - t^{2} cost + \int (2 t) cost dt$ $= - t^{2} cost + 2 {tsint - \int sint dt} = - t^{2} cost + 2 t sint + 2 cost$ $\int t^{2} \sin (3 t) dt =_{3 t = x} \int \frac{x^{2}}{9} sinx \frac{dx}{3} = \frac{1}{27} \int x^{2} sinx dx$ $= \frac{1}{27} {- x^{2} cosx + 2 x sinx + 2 cosx}$ $= \frac{1}{27} {- {(3 t)}^{2} \cos (3 t) + 6 t \sin (3 t) + 2 \cos (3 t)} \Rightarrow$ $v_{1} = \frac{1}{2} {- t^{2} cost + 2 tsint + 2 cost} - \frac{1}{54} {- {9 t}^{2} \cos (3 t) + 6 t \sin (3 t) + 2 \cos (3 t)}$ $v_{2} = \int \frac{w_{2}}{W} dt = \int t^{2} cost \cos (2 t) dt$ $= \frac{1}{2} \int t^{2} {\cos (3 t) + cost} dt = \frac{1}{2} \int t^{2} \cos (3 t) dt + \frac{1}{2} \int t^{2} cost dt$ $\int t^{2} cost dt = t^{2} sint - \int 2 t sint dt = t^{2} sint - 2 {- tcost + \int cost dt}$ $= t^{2} sint + 2 tcost - 2 cost$ $\int t^{2} \cos (3 t) dt =_{3 t = u} \int \frac{u^{2}}{9} \cos (u) \frac{du}{3} = \frac{1}{27} \int u^{2} cosu du$ $= \frac{1}{27} {u^{2} sinu + 2 u cosu - 2 cosu} = \frac{1}{27} {{9 t}^{2} \sin (3 t) + 6 t \cos (3 t) - 2 \cos (3 t)}$ $\Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2} and general solution is$ $y = y_{h} + y_{p}$
	Commented by Ar Brandon last updated on 14/Aug/20
	Thanks
	Commented by abdomathmax last updated on 14/Aug/20

	$you are welcome$
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