Solve the differential equation; x′′(t)+2x′(t)+x(t)=1+t (using the method of variation of parameters)


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Question Number 107995 by Ar Brandon last updated on 13/Aug/20

$Solve the differential equation;$ $x^{″} (t) + {2 x}^{'} (t) + x (t) = 1 + t$ $(using the method of variation of parameters)$
	Answered by Ar Brandon last updated on 13/Aug/20
	$x_(GH) : r^2 +2r+1=0 , (r+1)^2 =0, r_1 =−1=r_2 ⇒x_(GH) =(At+B)e^(−t) =Ate^(−t) +Be^(−t) By varying the parameters, Let x_P =A(t)te^(−t) +B(t)e^(−t) { ((A′(t)(te^(−t) )+B′(t)(e^(−t) )=0),(eqn(1))),((A′(t)(te^(−t) )′+B′(t)(e^(−t) )′=1+t),(eqn(2))) :} eqn(2)⇒A′(t)(e^(−t) −te^(−t) )−B′(t)(e^(−t) )=1+t ⇒−[A′(t)(te^(−t) )+B′(t)e^(−t) ]+A′(t)e^(−t) =1+t⇒A′(t)=e^t +te^t ⇒A(t)=∫(1+t)e^t dt=(1+t)e^t −e^t +k=te^t +k A(t) in eqn(1)⇒B′(t)=−A′(t)t⇒B(t)=−∫A′(t)tdt ⇒−B(t)=t∫A′(t)dt−∫{(dt/dt)∫A′(t)dt}dt ⇒B(t)=−t(te^t +k)+∫(te^t +k)dt ⇒B(t)=−t(te^t +k)+[te^t −e^t ]+kt+c ⇒B(t)=−t^2 e^t +te^t −e^t +c ⇒x_P =(te^t +k)te^(−t) +(−t^2 e^t +te^t −e^t +c)e^(−t) ⇒x_P =t^2 +kte^(−t) −t^2 +t−1+ce^(−t) =t−1+(kt+c)e^(−t) ⇒x_G =(At+B)e^(−t) +t−1+(kt+c)e^(−t) ⇒x_G =(αt+β)e^(−t) +t−1 , α=A+k , β=B+c$
	$x_{GH} : r^{2} + 2 r + 1 = 0, {(r + 1)}^{2} = 0, r_{1} = - 1 = r_{2}$ $\Rightarrow x_{GH} = (At + B) e^{- t} = {Ate}^{- t} + {Be}^{- t}$ $By varying the parameters,$ $Let x_{P} = A (t) {te}^{- t} + B (t) e^{- t}$ ${\begin{cases} A^{'} (t) ({te}^{- t}) + B^{'} (t) (e^{- t}) = 0 & eqn (1) \\ A^{'} (t) {({te}^{- t})}^{'} + B^{'} (t) {(e^{- t})}^{'} = 1 + t & eqn (2) \end{cases}$ $eqn (2) \Rightarrow A^{'} (t) (e^{- t} - {te}^{- t}) - B^{'} (t) (e^{- t}) = 1 + t$ $\Rightarrow - [A^{'} (t) ({te}^{- t}) + B^{'} (t) e^{- t}] + A^{'} (t) e^{- t} = 1 + t \Rightarrow A^{'} (t) = e^{t} + {te}^{t}$ $\Rightarrow A (t) = \int (1 + t) e^{t} dt = (1 + t) e^{t} - e^{t} + k = {te}^{t} + k$ $A (t) in eqn (1) \Rightarrow B^{'} (t) = - A^{'} (t) t \Rightarrow B (t) = - \int A^{'} (t) tdt$ $\Rightarrow - B (t) = t \int A^{'} (t) dt - \int {\frac{dt}{dt} \int A^{'} (t) dt} dt$ $\Rightarrow B (t) = - t ({te}^{t} + k) + \int ({te}^{t} + k) dt$ $\Rightarrow B (t) = - t ({te}^{t} + k) + [{te}^{t} - e^{t}] + kt + c$ $\Rightarrow B (t) = - t^{2} e^{t} + {te}^{t} - e^{t} + c$ $\Rightarrow x_{P} = ({te}^{t} + k) {te}^{- t} + (- t^{2} e^{t} + {te}^{t} - e^{t} + c) e^{- t}$ $\Rightarrow x_{P} = t^{2} + {kte}^{- t} - t^{2} + t - 1 + {ce}^{- t} = t - 1 + (kt + c) e^{- t}$ $\Rightarrow x_{G} = (At + B) e^{- t} + t - 1 + (kt + c) e^{- t}$ $\Rightarrow x_{G} = (α t + β) e^{- t} + t - 1, α = A + k, β = B + c$
	Answered by Aziztisffola last updated on 13/Aug/20

	$x^{″} (t) + {2 x}^{'} (t) + x (t) = 0 \Rightarrow r^{2} + 2 r + 1 = 0$ ${(r + 1)}^{2} = 0 \Rightarrow r = - 1$ $x_{h} (t) = (α t + β) e^{- t} = α {te}^{- t} + β e^{- t}$ $x_{p} (t) = u_{1} {te}^{- t} + u_{2} e^{- t}$ $w = \| \begin{matrix} {te}^{- t} & e^{- t} \\ (1 - t) e^{- t} & - e^{- t} \end{matrix} \| = - {te}^{- 2 t} - (1 - t) e^{- 2 t}$ $= (- t - 1 + t) e^{- 2 t} = - e^{- 2 t}$ $w_{1} = \| \begin{matrix} 0 & e^{- t} \\ 1 + t & - e^{- t} \end{matrix} \| = - e^{- t} (1 + t)$ $w_{2} = \| \begin{matrix} {te}^{- t} & 0 \\ (1 - t) e^{- t} & 1 + t \end{matrix} \| = (1 + t) {te}^{- t}$ $u_{1} = \int \frac{w_{1}}{w} dt = \int (1 + t) e^{t} dt$ $u_{2} = \int \frac{w_{2}}{w} dt = - \int (1 + t) {te}^{t} dt$ $x_{p} (t) = {te}^{- t} \int (1 + t) e^{t} dt - e^{- t} \int (1 + t) {te}^{t} dt$ $x (t) = x_{h} (t) + x_{p} (t)$
	Commented by Ar Brandon last updated on 13/Aug/20
	Thanks
	Answered by mathmax by abdo last updated on 14/Aug/20

	$laplace method$ $e \Rightarrow L (y^{^{″}}) + 2 L (y^{^{'}}) + L (y) = L (1 + t) \Rightarrow$ $t^{2} L (y) - ty (o) - y^{^{'}} (0) + 2 (t L (y) - y (0)) + L (y) = L (1 + t)$ $\Rightarrow (t^{2} + 2 t + 1) L (y) = ty (o) + 2 y (o) + y^{^{'}} (0) + L (1 + t) \Rightarrow$ $(t^{2} + 2 t + 1) L (y) = (t + 2) y (o) + y^{^{'}} (0) + L (1 + t) but$ $L (1 + t) = \int_{0}^{\infty} (1 + x) e^{- tx} dx = \int_{0}^{\infty} e^{- tx} dx + \int_{0}^{\infty} {xe}^{- tx} dx$ $= {[- \frac{1}{t} e^{- tx}]}_{x = 0}^{\infty} + {[- \frac{x}{t} e^{- tx}]}_{x = 0}^{\infty} + \frac{1}{t} \int_{0}^{\infty} e^{- tx} dx$ $= \frac{1}{t} + \frac{1}{t} {[- \frac{1}{t} e^{- tx}]}_{x = 0}^{\infty} = \frac{1}{t} + \frac{1}{t^{2}}$ $e \Rightarrow (t^{2} + 2 t + 1) L (y) = y (0) (t + 2) + y^{^{'}} (0) + \frac{1}{t} + \frac{1}{t^{2}} \Rightarrow$ $\Rightarrow L (y) = y (0) . \frac{t + 2}{{(t + 1)}^{2}} + y^{^{'}} (0) \frac{1}{{(t + 1)}^{2}} + \frac{1}{t {(t + 1)}^{2}} + \frac{1}{t^{2} {(t + 1)}^{2}} \Rightarrow$ $y (t) = y (0) L^{- 1} (\frac{t + 2}{{(t + 2)}^{2}}) + y^{^{'}} (0) L^{- 1} (\frac{1}{{(t + 1)}^{2}}) + L^{- 1} (\frac{1}{t {(t + 1)}^{2}})$ $+ L^{- 1} (\frac{1}{t^{2} {(t + 1)}^{2}}) rest decompisition . . . be continued . . .$
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