Solve by Method of variation parameter (d^2 y/dx^2 )−3(dy/dx)+2y=sinx M.m


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Question Number 175213 by Mastermind last updated on 23/Aug/22

$Solve by Method of variation parameter$ $\frac{d^{2} y}{{dx}^{2}} - 3 \frac{dy}{dx} + 2 y = sinx$ $M . m$
	Answered by Ar Brandon last updated on 23/Aug/22
	$(d^2 y/dx^2 )−3(dy/dx)+2y=sinx (h.e.)→r^2 −3r+2=0, r=1, r=2 y_h =αe^x +βe^(2x) ⇒y_p =α(x)e^x +β(x)e^(2x) { ((α′(x)e^x +β′(x)e^(2x) =0)),((α′(x)(e^x )′+β′(x)(e^(2x) )′=sinx)) :}⇒ { ((α′(x)e^x +β′(x)e^(2x) =0)),((α′(x)e^x +2β′(x)e^(2x) =sinx)) :} ⇒W= determinant ((e^x ,e^(2x) ),(e^x ,(2e^(2x) )))=2e^(3x) −e^(3x) =e^(3x) ⇒W_α = determinant ((0,e^(2x) ),((sinx),(2e^(2x) )))=−e^(2x) sinx ; W_β = determinant ((e^x ,0),(e^x ,(sinx)))=e^x sinx α(x)=∫(W_α /W)dx=−∫e^(−x) sinxdx=(e^(−x) /2)(sinx+cosx) β(x)=∫(W_β /W)dx=∫e^(−2x) sinxdx=−(1/2)e^(−2x) sinx+(1/2)∫e^(−2x) cosxdx =−(1/2)e^(−2x) sinx+(1/2){−(1/2)e^(−2x) cosx−(1/2)∫e^(−2x) sinxdx} =(4/5)(−(1/2)e^(−2x) sinx−(1/4)e^(−2x) cosx)=−(e^(−2x) /5)(2sinx+cosx) ⇒y_p =(1/2)(sinx+cosx)−(1/5)(2sinx+cosx)=(1/(10))sinx+(3/(10))cosx Y=y_h +y_p =αe^x +βe^(2x) +(1/(10))sinx+(3/(10))cosx$
	$\frac{d^{2} y}{{d x}^{2}} - 3 \frac{d y}{d x} + 2 y = \sin x$ $(h . e .) \to r^{2} - 3 r + 2 = 0, r = 1, r = 2$ $y_{h} = α e^{x} + β e^{2 x} \Rightarrow y_{p} = α (x) e^{x} + β (x) e^{2 x}$ ${\begin{cases} α^{'} (x) e^{x} + β^{'} (x) e^{2 x} = 0 \\ α^{'} (x) {(e^{x})}^{'} + β^{'} (x) {(e^{2 x})}^{'} = \sin x \end{cases} \Rightarrow {\begin{cases} α^{'} (x) e^{x} + β^{'} (x) e^{2 x} = 0 \\ α^{'} (x) e^{x} + 2 β^{'} (x) e^{2 x} = \sin x \end{cases}$ $\Rightarrow W = \| \begin{matrix} e^{x} & e^{2 x} \\ e^{x} & 2 e^{2 x} \end{matrix} \| = 2 e^{3 x} - e^{3 x} = e^{3 x}$ $\Rightarrow W_{α} = \| \begin{matrix} 0 & e^{2 x} \\ \sin x & 2 e^{2 x} \end{matrix} \| = - e^{2 x} \sin x; W_{β} = \| \begin{matrix} e^{x} & 0 \\ e^{x} & \sin x \end{matrix} \| = e^{x} \sin x$ $α (x) = \int \frac{W_{α}}{W} d x = - \int e^{- x} \sin x d x = \frac{e^{- x}}{2} (\sin x + \cos x)$ $β (x) = \int \frac{W_{β}}{W} d x = \int e^{- 2 x} \sin x d x = - \frac{1}{2} e^{- 2 x} \sin x + \frac{1}{2} \int e^{- 2 x} \cos x d x$ $= - \frac{1}{2} e^{- 2 x} \sin x + \frac{1}{2} {- \frac{1}{2} e^{- 2 x} \cos x - \frac{1}{2} \int e^{- 2 x} \sin x d x}$ $= \frac{4}{5} (- \frac{1}{2} e^{- 2 x} \sin x - \frac{1}{4} e^{- 2 x} \cos x) = - \frac{e^{- 2 x}}{5} (2 \sin x + \cos x)$ $\Rightarrow y_{p} = \frac{1}{2} (\sin x + \cos x) - \frac{1}{5} (2 \sin x + \cos x) = \frac{1}{10} \sin x + \frac{3}{10} \cos x$ $Y = y_{h} + y_{p} = α e^{x} + β e^{2 x} + \frac{1}{10} \sin x + \frac{3}{10} \cos x$
	Commented by Tawa11 last updated on 25/Aug/22

	$Great Sir .$
	Answered by Ar Brandon last updated on 23/Aug/22
	$y_h =αe^x +βe^(2x) y_p =acosx+bsinx (−acosx−bsinx)−3(−asinx+bcosx) +2(acosx+bsinx)=sinx ⇒(a−3b)cosx+(3a+b)sinx=sinx { ((a−3b=0)),((3a+b=1)) :} ⇒10a=3⇒a=(3/(10)) , b=(1/(10)) ⇒y_p =(3/(10))cosx+(1/(10))sinx Y=y_h +y_p =αe^x +βe^(2x) +(3/(10))cosx+(1/(10))sinx$
	$y_{h} = α e^{x} + β e^{2 x}$ $y_{p} = a \cos x + b \sin x$ $(- a \cos x - b \sin x) - 3 (- a \sin x + b \cos x)$ $+ 2 (a \cos x + b \sin x) = \sin x$ $\Rightarrow (a - 3 b) \cos x + (3 a + b) \sin x = \sin x$ ${\begin{cases} a - 3 b = 0 \\ 3 a + b = 1 \end{cases} \Rightarrow 10 a = 3 \Rightarrow a = \frac{3}{10}, b = \frac{1}{10}$ $\Rightarrow y_{p} = \frac{3}{10} \cos x + \frac{1}{10} \sin x$ $Y = y_{h} + y_{p} = α e^{x} + β e^{2 x} + \frac{3}{10} \cos x + \frac{1}{10} \sin x$
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