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Question Number 100766 by john santu last updated on 28/Jun/20

	Answered by Rio Michael last updated on 28/Jun/20

	$A E : m^{2} + 1 = 0 \Rightarrow m = \pm i$ $y_{c} = (A \cos x + B \sin x)$ $let y_{p} = λ e^{2 x} \Rightarrow y^{'} = 2 λ e^{2 x} and y^{″} = 4 λ e^{2 x}$ $\Rightarrow 4 λ e^{2 x} + λ e^{2 x} = e^{2 x} [$ $\Rightarrow 5 λ = 1 \Leftrightarrow λ = \frac{1}{5}$ $\Rightarrow y_{p} = \frac{e^{2 x}}{5}$ $y_{g} = y_{c} + y_{p} \Rightarrow y_{g} = A \cos x + B \sin x + \frac{e^{2 x}}{5}$
	Answered by mathmax by abdo last updated on 28/Jun/20
	$y^(′′) +y =e^(2x) (he)→y^(′′) +y =0 →r^2 +1=0⇒r =+^− i ⇒y_h =acosx +bsinx =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((cosx sinx)),((−sinx cosx)))=cos^2 x +sin^2 x =1 W_1 = determinant (((0 sinx)),((e^(2x) cosx)))=−e^(2x ) sinx W_2 = determinant (((cosx 0)),((−sinx e^(2x) )))=e^(2x) cosx v_1 =∫ (w_1 /w)dx =−∫ e^(2x) cosx dx =−Re(∫ e^(2x +ix) dx) and ∫ e^((2+i)x) ex =(1/(2+i)) e^((2+i)x) =(e^(2x) /(2+i))(cosx +isinx) =(1/5) e^(2x) (2−i)(cosx +isinx) =(e^(2x) /5){2cosx+2isinx −icosx+sinx} ⇒ v_1 =−(e^(2x) /5){2cosx+sinx} v_2 =∫ (w_2 /w)dx =∫e^(2x) cosx dx =(e^(2x) /5){2cosx +sinx} ⇒y_p =u_1 v_1 +u_2 v_2 =−(e^(2x) /5){2cos^2 x +cosx sinx} +(e^(2x) /5){ 2sinx cosx +sin^2 x} =(e^(2x) /5){2sinx cosx+sin^2 x−2cos^2 x−cosx sinx} =(e^(2x) /5){sinx cosx +1−3cos^2 x} =(e^(2x) /5){(1/2)sin(2x)+1−3×((1+cos(2x))/2)} =(e^(2x) /5){(1/2)sin(2x)−(1/2)−(3/2)cos(2x)} =(e^(2x) /(10)){ sin(2x)−3cos(2x)−1} y =y_h +y_p =acosx +bsinx +(e^(2x) /(10)){ sin(2x)−3cos(2x)−1}$
	$y^{^{″}} + y = e^{2 x} (he) \to y^{^{″}} + y = 0 \to r^{2} + 1 = 0 \Rightarrow r = \bar{+} i \Rightarrow y_{h} = acosx + bsinx$ $= {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} cosx sinx \\ - sinx cosx \end{matrix} \| = \cos^{2} x + \sin^{2} x = 1$ $W_{1} = \| \begin{matrix} 0 sinx \\ e^{2 x} cosx \end{matrix} \| = - e^{2 x} sinx$ $W_{2} = \| \begin{matrix} cosx 0 \\ - sinx e^{2 x} \end{matrix} \| = e^{2 x} cosx$ $v_{1} = \int \frac{w_{1}}{w} dx = - \int e^{2 x} cosx dx = - Re (\int e^{2 x + ix} dx) and$ $\int e^{(2 + i) x} ex = \frac{1}{2 + i} e^{(2 + i) x} = \frac{e^{2 x}}{2 + i} (cosx + isinx)$ $= \frac{1}{5} e^{2 x} (2 - i) (cosx + isinx) = \frac{e^{2 x}}{5} {2 cosx + 2 isinx - icosx + sinx} \Rightarrow$ $v_{1} = - \frac{e^{2 x}}{5} {2 cosx + sinx}$ $v_{2} = \int \frac{w_{2}}{w} dx = \int e^{2 x} cosx dx = \frac{e^{2 x}}{5} {2 cosx + sinx} \Rightarrow y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $= - \frac{e^{2 x}}{5} {{2 \cos}^{2} x + cosx sinx} + \frac{e^{2 x}}{5} {2 sinx cosx + \sin^{2} x}$ $= \frac{e^{2 x}}{5} {2 sinx cosx + \sin^{2} x - {2 \cos}^{2} x - cosx sinx}$ $= \frac{e^{2 x}}{5} {sinx cosx + 1 - {3 \cos}^{2} x} = \frac{e^{2 x}}{5} {\frac{1}{2} \sin (2 x) + 1 - 3 \times \frac{1 + \cos (2 x)}{2}}$ $= \frac{e^{2 x}}{5} {\frac{1}{2} \sin (2 x) - \frac{1}{2} - \frac{3}{2} \cos (2 x)} = \frac{e^{2 x}}{10} {\sin (2 x) - 3 \cos (2 x) - 1}$ $y = y_{h} + y_{p} = acosx + bsinx + \frac{e^{2 x}}{10} {\sin (2 x) - 3 \cos (2 x) - 1}$
	Answered by mathmax by abdo last updated on 28/Jun/20

	$Laplace method y^{^{″}} + y = e^{2 x} \Rightarrow L (y^{^{″}}) + L (y) = L (e^{2 x}) \Rightarrow$ $x^{2} L (y) - xy (0) - y^{^{'}} (0) + L (y) = L (e^{2 x}) \Rightarrow (x^{2} + 1) L (y) = xy (0) + y^{^{'}} (0) + L (e^{2 x})$ $we have L (e^{2 x}) = \int_{0}^{\infty} e^{2 t} e^{- xt} dt = \int_{0}^{\infty} e^{(2 - x) t} dt = {[\frac{1}{2 - x} e^{(2 - x) t}]}_{0}^{+ \infty}$ $= - \frac{1}{2 - x} = \frac{1}{x - 2} \Rightarrow (x^{2} + 1) L (y) = \frac{1}{x - 2} + xy (o) + y^{^{'}} (0) \Rightarrow$ $L (y) = \frac{1}{(x - 2) (x^{2} + 1)} + \frac{x}{x^{2} + 1} y (o) + \frac{y^{^{'}} (0)}{x^{2} + 1} \Rightarrow$ $y (x) = L^{- 1} (\frac{1}{(x - 2) (x^{2} + 1)}) + y (o) L^{- 1} (\frac{x}{x^{2} + 1}) + y^{^{'}} (o) L^{- 1} (\frac{1}{x^{2} + 1})$ $we have L^{- 1} (\frac{x}{x^{2} + 1}) = cosx and L^{- 1} (\frac{1}{x^{2} + 1}) = sinx$ $f (x) = \frac{1}{(x - 2) (x^{2} + 1)} = \frac{1}{(x - 2) (x - i) (x + i)} = \frac{a}{x - 2} + \frac{b}{x - i} + \frac{c}{x + i}$ $L^{- 1} (f (x)) = {ae}^{2 x} + b e^{ix} + {ce}^{- ix} \to {ae}^{2 x} + α cosx + β sinx \Rightarrow$ $a = \frac{1}{5} \Rightarrow y (x) = \frac{1}{5} e^{2 x} + A cosx + B sinx$
	Answered by john santu last updated on 28/Jun/20

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