y′′−2ay′+(1+a^2 )y=te^(at) +sint


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Question Number 113444 by Ar Brandon last updated on 13/Sep/20

$y^{″} - {2 ay}^{'} + (1 + a^{2}) y = {te}^{at} + sint$
	Answered by Ar Brandon last updated on 13/Sep/20
	$y′′−2ay′+(1+a^2 )y=te^(at) +sint (h)→r^2 −2ar+1+a^2 =0⇒r=((2a±(√(4a^2 −4(1+a^2 ))))/2)=a±i y_h =e^(at) (Acost+Bsint) By varying parameters let; y_p =A(t)e^(at) cost+B(t)e^(at) sint ⇒y_p =au_1 +bu_2 { ((a′u_1 +b′u_2 =0),(...(1))),((a′u_1 ′+b′u_2 ′=te^(at) +sint),(...(2))) :} Finding a′ and b′ using Crammer′s method; W(u_1 ,u_2 )= determinant ((u_1 ,u_2 ),((u_1 ′),(u_2 ′)))= determinant (((e^(at) cost),(e^(at) sint)),((ae^(at) cost−e^(at) sint),(e^(at) cost+ae^(at) sint))) =e^(2at) [(cos^2 t+asintcost)−(asintcost−sin^2 t)]=e^(2at) w_1 = determinant ((0,(e^(at) sint)),((te^(at) +sint),(e^(at) cost+ae^(at) sint)))=−(te^(2at) sint+e^(at) sin^2 t) w_2 = determinant (((e^(at) cost),0),((ae^(at) cost−e^(at) sint),(te^(at) +sint)))=te^(2at) cost+e^(at) sintcost a′=(w_1 /W)⇒a=∫(w_1 /W)dt , b′=(w_2 /W)⇒b=∫(w_2 /W)dt ⇒A(t)=−∫((te^(2at) sint+e^(at) sin^2 t)/e^(2at) )dt, B(t)=∫((te^(2at) cost+e^(at) sintcost)/e^(2at) )dt A(t)=−∫(tsint+e^(−at) sin^2 t)dt ∫tsintdt=−tcost+∫costdt=sint−tcost ∫e^(−at) sin^2 tdt=−((sin^2 t)/a)∙e^(−at) +(2/a)∫sintcost∙e^(−at) dt =−((sin^2 t)/a)∙e^(−at) +(1/a)∫sin(2t)e^(−at) dt =−((sin^2 t)/a)∙e^(−at) +(1/a){−((sin(2t))/a)e^(−at) +(2/a)∫cos(2t)e^(−at) dt} =−((sin^2 t)/a)∙e^(−at) −((sin(2t))/a^2 )e^(−at) +(2/a^2 )[−((cos(2t))/a)e^(−at) −(2/a)∫sin(2t)e^(−at) dt] =−((sin^2 t)/a)∙e^(−at) −((sin(2t))/a^2 )e^(−at) −((2cos(2t))/a^3 )e^(−at) −(4/a^3 )∫sin(2t)e^(−at) dt =−((sin^2 t)/a)∙e^(−at) +(a^3 /(a^2 +4))[−((sin(2t))/a^2 )e^(−at) −((2cos(2t))/a^3 )e^(−at) ]+C ⇒A(t)=tcost−sint+((sin^2 t)/a)∙e^(−at) +(a^3 /(a^2 +4))[((sin(2t))/a^2 )e^(−at) +((2cos(2t))/a^3 )e^(−at) ]+C B(t)=∫((te^(2at) cost+e^(at) sintcost)/e^(2at) )dt=∫(tcost+e^(−at) sintcost)dt ∫tcostdt=tsint−∫sintdt=tsint+cost ∫e^(−at) sintcostdt=(1/2)∫e^(−at) sin(2t)dt =(1/2)∙(a^4 /(a^2 +4))[−((sin(2t))/a^2 )e^(−at) −((2cos(2t))/a^3 )e^(−at) ]+K ⇒B(t)=tsint+cost+(1/2)∙(a^4 /(a^2 +4))[−((sin(2t))/a^2 )e^(−at) −((2cos(2t))/a^3 )e^(−at) ]+K Or y_p =A(t)e^(at) cost+B(t)e^(at) sint et y_G =y_h +y_p y_G =e^(at) (Acost+Bsint)+e^(at) tcos^2 t−e^(at) sintcost+((sin^2 tcost)/a)+((a^3 cost)/(a^2 +4))[((sin(2t))/a^2 )+((2cos(2t))/a^3 )]+Ce^(at) cost e^(at) tsintcost+e^(at) cos^2 t+(1/2)∙((a^4 cost)/(a^2 +4))[−((sin(2t))/a^2 )−((2cos(2t))/a^3 )]+Ke^(at) cost$
	$y^{″} - {2 ay}^{'} + (1 + a^{2}) y = {te}^{at} + sint$ $(h) \to r^{2} - 2 ar + 1 + a^{2} = 0 \Rightarrow r = \frac{2 a \pm \sqrt{{4 a}^{2} - 4 (1 + a^{2})}}{2} = a \pm i$ $y_{h} = e^{at} (Acost + Bsint)$ $By varying parameters let; y_{p} = A (t) e^{at} cost + B (t) e^{at} sint$ $\Rightarrow y_{p} = {au}_{1} + {bu}_{2}$ ${\begin{cases} a^{'} u_{1} + b^{'} u_{2} = 0 & . . . (1) \\ a^{'} u_{1}^{'} + b^{'} u_{2}^{'} = {te}^{at} + sint & . . . (2) \end{cases}$ $Finding a^{'} and b^{'} using {Crammer}^{'} s method;$ $W (u_{1}, u_{2}) = \| \begin{matrix} u_{1} & u_{2} \\ u_{1}^{'} & u_{2}^{'} \end{matrix} \| = \| \begin{matrix} e^{at} cost & e^{at} sint \\ {ae}^{at} cost - e^{at} sint & e^{at} cost + {ae}^{at} sint \end{matrix} \|$ $= e^{2 at} [(\cos^{2} t + asintcost) - (asintcost - \sin^{2} t)] = e^{2 at}$ $w_{1} = \| \begin{matrix} 0 & e^{at} sint \\ {te}^{at} + sint & e^{at} cost + {ae}^{at} sint \end{matrix} \| = - ({te}^{2 at} sint + e^{at} \sin^{2} t)$ $w_{2} = \| \begin{matrix} e^{at} cost & 0 \\ {ae}^{at} cost - e^{at} sint & {te}^{at} + sint \end{matrix} \| = {te}^{2 at} cost + e^{at} sintcost$ $a^{'} = \frac{w_{1}}{W} \Rightarrow a = \int \frac{w_{1}}{W} dt, b^{'} = \frac{w_{2}}{W} \Rightarrow b = \int \frac{w_{2}}{W} dt$ $\Rightarrow A (t) = - \int \frac{{te}^{2 at} sint + e^{at} \sin^{2} t}{e^{2 at}} dt, B (t) = \int \frac{{te}^{2 at} cost + e^{at} sintcost}{e^{2 at}} dt$ $A (t) = - \int (tsint + e^{- at} \sin^{2} t) dt$ $\int tsintdt = - tcost + \int costdt = sint - tcost$ $\int e^{- at} \sin^{2} tdt = - \frac{\sin^{2} t}{a} \cdot e^{- at} + \frac{2}{a} \int sintcost \cdot e^{- at} dt$ $= - \frac{\sin^{2} t}{a} \cdot e^{- at} + \frac{1}{a} \int \sin (2 t) e^{- at} dt$ $= - \frac{\sin^{2} t}{a} \cdot e^{- at} + \frac{1}{a} {- \frac{\sin (2 t)}{a} e^{- at} + \frac{2}{a} \int \cos (2 t) e^{- at} dt}$ $= - \frac{\sin^{2} t}{a} \cdot e^{- at} - \frac{\sin (2 t)}{a^{2}} e^{- at} + \frac{2}{a^{2}} [- \frac{\cos (2 t)}{a} e^{- at} - \frac{2}{a} \int \sin (2 t) e^{- at} dt]$ $= - \frac{\sin^{2} t}{a} \cdot e^{- at} - \frac{\sin (2 t)}{a^{2}} e^{- at} - \frac{2 \cos (2 t)}{a^{3}} e^{- at} - \frac{4}{a^{3}} \int \sin (2 t) e^{- at} dt$ $= - \frac{\sin^{2} t}{a} \cdot e^{- at} + \frac{a^{3}}{a^{2} + 4} [- \frac{\sin (2 t)}{a^{2}} e^{- at} - \frac{2 \cos (2 t)}{a^{3}} e^{- at}] + C$ $\Rightarrow A (t) = tcost - sint + \frac{\sin^{2} t}{a} \cdot e^{- at} + \frac{a^{3}}{a^{2} + 4} [\frac{\sin (2 t)}{a^{2}} e^{- at} + \frac{2 \cos (2 t)}{a^{3}} e^{- at}] + C$ $B (t) = \int \frac{{te}^{2 at} cost + e^{at} sintcost}{e^{2 at}} dt = \int (tcost + e^{- at} sintcost) dt$ $\int tcostdt = tsint - \int sintdt = tsint + cost$ $\int e^{- at} sintcostdt = \frac{1}{2} \int e^{- at} \sin (2 t) dt$ $= \frac{1}{2} \cdot \frac{a^{4}}{a^{2} + 4} [- \frac{\sin (2 t)}{a^{2}} e^{- at} - \frac{2 \cos (2 t)}{a^{3}} e^{- at}] + K$ $\Rightarrow B (t) = tsint + cost + \frac{1}{2} \cdot \frac{a^{4}}{a^{2} + 4} [- \frac{\sin (2 t)}{a^{2}} e^{- at} - \frac{2 \cos (2 t)}{a^{3}} e^{- at}] + K$ $Or y_{p} = A (t) e^{at} cost + B (t) e^{at} sint et y_{G} = y_{h} + y_{p}$ $y_{G} = e^{at} (Acost + Bsint) + e^{at} {tcos}^{2} t - e^{at} sintcost + \frac{\sin^{2} tcost}{a} + \frac{a^{3} cost}{a^{2} + 4} [\frac{\sin (2 t)}{a^{2}} + \frac{2 \cos (2 t)}{a^{3}}] + {Ce}^{at} cost$ $e^{at} tsintcost + e^{at} \cos^{2} t + \frac{1}{2} \cdot \frac{a^{4} cost}{a^{2} + 4} [- \frac{\sin (2 t)}{a^{2}} - \frac{2 \cos (2 t)}{a^{3}}] + {Ke}^{at} cost$
	Commented by bemath last updated on 13/Sep/20

	$a m a z i n g$
	Commented by bobhans last updated on 13/Sep/20

	$super great . . . .$
	Commented by Ar Brandon last updated on 13/Sep/20
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