(d^2 y/dx) − 3 (dy/dx) + 2y = e^(4t) , y(0) = 1, y′(0) = 0 solve with Laplace Transform!


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Question Number 126620 by fajri last updated on 22/Dec/20

$\frac{d^{2} y}{d x} - 3 \frac{d y}{d x} + 2 y = e^{4 t}, y (0) = 1, y^{'} (0) = 0$ $s o l v e w i t h L a p l a c e T r a n s f o r m!$
	Answered by Olaf last updated on 22/Dec/20

	$\frac{d^{2} y}{{d x}^{2}} - 3 \frac{d y}{d x} + 2 y = e^{4 t}$ $[p^{2} L (y) - p y (0^{-}) - y^{'} (0^{-})] - 3 [p L (y) - y (0^{-})]$ $+ 2 L (y) = \frac{1}{p - 4}$ $(p^{2} - 3 p + 2) L (y) - p + 3 = \frac{1}{p - 4}$ $(p - 1) (p - 2) L (y) = \frac{1}{p - 4} + p - 3$ $L (y) = \frac{1 + (p - 3) (p - 4)}{(p - 1) (p - 2) (p - 4)}$ $L (y) = \frac{7 / 3}{p - 1} - \frac{3 / 2}{p - 2} + \frac{1 / 6}{p - 4}$ $L^{- 1} o L (y) = y = \frac{7}{3} e^{t} - \frac{3}{2} e^{2 t} + \frac{1}{6} e^{4 t}$
	Commented by fajri last updated on 22/Dec/20

	$t h a n k s S i r, L i k e i t!!$
	Answered by mathmax by abdo last updated on 22/Dec/20
	$wronskien method y^(′′) −3y^′ +2y=e^(4x) with y(o)=1 and y^′ (0)=0 h)→r^2 −3r +2=0→Δ=1 ⇒r_1 =((3+1)/2)=2 and r_2 =((3−1)/2)=1 ⇒ y_h =ae^x +be^(2x) =au_1 +bu_2 W(u_1 ,u_2 )= determinant (((e^x e^(2x) )),((e^x 2e^(2x) )))=2e^(3x) −e^(3x) =e^(3x) ≠0 W_1 = determinant (((o e^(2x) )),((e^(4x) 2e^(2x) )))=−e^(6x) W_2 = determinant (((e^x o)),((e^x e^(4x) )))=e^(5x) v_1 =∫ (W_1 /W)dx =∫ ((−e^(6x) )/e^(3x) )dx =−∫ e^(3x) dx =−(1/3)e^(3x) v_2 =∫ (W_2 /W)dx =∫ (e^(5x) /e^(3x) )dx =∫ e^(2x) dx =(1/2)e^(2x) ⇒ y_p =u_1 v_1 +u_2 v_2 =e^x (−(1/3)e^(3x) )+e^(2x) ((1/2)e^(2x) )=−(1/3)e^(4x) +(1/2)e^(4x) =(1/6)e^(4x) ⇒y(x)=ae^x +be^(2x) +(1/6)e^(4x) y(0)=1 ⇒a+b+(1/6)=1 ⇒a+b=1−(1/6)=(5/6) y^′ (x)=ae^x +2b e^(2x) +(2/3)e^(4x) y^′ (o)=0 ⇒a+2b +(2/3)=0 ⇒a+2b=−(2/3) we get the system { ((a+b=(5/6) ⇒ b=−(2/3)−(5/6)=−(9/6)=−(3/2))),((a+2b=−(2/3))) :} a=(5/6)+(3/2)=((5+9)/6)=((14)/6)=(7/3) ⇒★y(x)=(7/3)e^x −(3/2)e^(2x) +(1/6)e^(4x) ★$
	$wronskien method y^{^{″}} - {3 y}^{^{'}} + 2 y = e^{4 x} with y (o) = 1 and y^{^{'}} (0) = 0$ $h) \to r^{2} - 3 r + 2 = 0 \to Δ = 1 \Rightarrow r_{1} = \frac{3 + 1}{2} = 2 and r_{2} = \frac{3 - 1}{2} = 1 \Rightarrow$ $y_{h} = {ae}^{x} + {be}^{2 x} = {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{x} e^{2 x} \\ e^{x} {2 e}^{2 x} \end{matrix} \| = {2 e}^{3 x} - e^{3 x} = e^{3 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{2 x} \\ e^{4 x} {2 e}^{2 x} \end{matrix} \| = - e^{6 x}$ $W_{2} = \| \begin{matrix} e^{x} o \\ e^{x} e^{4 x} \end{matrix} \| = e^{5 x}$ $v_{1} = \int \frac{W_{1}}{W} dx = \int \frac{- e^{6 x}}{e^{3 x}} dx = - \int e^{3 x} dx = - \frac{1}{3} e^{3 x}$ $v_{2} = \int \frac{W_{2}}{W} dx = \int \frac{e^{5 x}}{e^{3 x}} dx = \int e^{2 x} dx = \frac{1}{2} e^{2 x} \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = e^{x} (- \frac{1}{3} e^{3 x}) + e^{2 x} (\frac{1}{2} e^{2 x}) = - \frac{1}{3} e^{4 x} + \frac{1}{2} e^{4 x}$ $= \frac{1}{6} e^{4 x} \Rightarrow y (x) = {ae}^{x} + {be}^{2 x} + \frac{1}{6} e^{4 x}$ $y (0) = 1 \Rightarrow a + b + \frac{1}{6} = 1 \Rightarrow a + b = 1 - \frac{1}{6} = \frac{5}{6}$ $y^{^{'}} (x) = {ae}^{x} + 2 b e^{2 x} + \frac{2}{3} e^{4 x}$ $y^{^{'}} (o) = 0 \Rightarrow a + 2 b + \frac{2}{3} = 0 \Rightarrow a + 2 b = - \frac{2}{3} we get the system$ ${\begin{cases} a + b = \frac{5}{6} \Rightarrow b = - \frac{2}{3} - \frac{5}{6} = - \frac{9}{6} = - \frac{3}{2} \\ a + 2 b = - \frac{2}{3} \end{cases}$ $a = \frac{5}{6} + \frac{3}{2} = \frac{5 + 9}{6} = \frac{14}{6} = \frac{7}{3} \Rightarrow ★ y (x) = \frac{7}{3} e^{x} - \frac{3}{2} e^{2 x} + \frac{1}{6} e^{4 x} ★$
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