(D^2 −6D+9)y = (e^(3x) /x^2 )


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Question Number 115721 by bemath last updated on 28/Sep/20

$(D^{2} - 6 D + 9) y = \frac{e^{3 x}}{x^{2}}$
Commented by mohammad17 last updated on 28/Sep/20

$m^{2} - 6 m + 9 = 0 \Rightarrow {(m - 3)}^{2} = 0 \Rightarrow m_{1} = m_{2} = 3$ $Y c = c_{1} e^{3 x} + c_{2} {x e}^{3 x}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{3 x} {x e}^{3 x} \\ 3 e^{3 x} 3 {x e}^{3 x} + e^{3 x} \end{matrix} \| = e^{6 x} \neq 0$ $W_{1} = \| \begin{matrix} 0 {x e}^{3 x} \\ \frac{e^{3 x}}{x^{2}} 3 {x e}^{3 x} + e^{3 x} \end{matrix} \| = - \frac{e^{6 x}}{x}$ $W_{2} = \| \begin{matrix} e^{3 x} 0 \\ 3 e^{3 x} \frac{e^{3 x}}{x^{2}} \end{matrix} \| = \frac{e^{6 x}}{x^{2}}$ $V_{1} = \int \frac{W_{1}}{W} d x = \int - \frac{1}{x^{2}} d x = \frac{1}{x}$ $V_{2} = \int \frac{W_{2}}{W} d x = \int \frac{1}{x^{2}} d x = - \frac{1}{x}$ $Y p = V_{1} U_{1} + V_{2} U_{2}$ $Y p = \frac{e^{3 x}}{x} - e^{3 x}$ $Y = Y c + Y p$ $Y = c_{1} e^{3 x} + c_{2} {x e}^{3 x} + \frac{e^{3 x}}{x} - e^{3 x}$ $(m . o)$
Commented by mohammad17 last updated on 28/Sep/20

$Y c = c_{1} e^{3 x} + c_{2} {x e}^{3 x}$ $Y p = e^{a x} . \frac{1}{f (D + a)} f (x)$ $Y p = e^{3 x} \frac{1}{{(D + 3)}^{2} - 6 (D + 3) + 9} . (\frac{1}{x^{2}})$ $Y p = e^{3 x} \frac{1}{D^{2}} (\frac{1}{x^{2}}) \Rightarrow Y p = - e^{3 x} \frac{1}{D} (\frac{1}{x})$ $Y p = - e^{3 x} l n x$ $Y = Y c + Y p$ $Y = c_{1} e^{3 x} + c_{2} {x e}^{3 x} - e^{3 x} l n x$
Commented by bemath last updated on 28/Sep/20

$g a v e k u d o s a l l m a s t e r$
Commented by mohammad17 last updated on 28/Sep/20

$t h a n k y o u s i r$
	Answered by john santu last updated on 28/Sep/20
	$let y = e^(3x) .q → { ((Dy=e^(3x) (q′+3q))),((D^2 y=e^(3x) (q′′+6q′+9q))) :} we obtain e^(3x) (q′′+6q′+9q)−6.e^(3x) (q′+3q)+9e^(3x) q=(e^(3x) /x^2 ) which simplifies to q′′ = (1/x^2 ) → ((d(dq))/dx) = (1/x^2 ) ∫ ((d(dq))/dx) = ∫ (1/x^2 ) ⇒ (dq/dx) = −(1/x) +C_1 ∫ dq = ∫ (−(1/x) +C_1 ) dx →q = −ln ∣x∣ + C_1 x+C_2 ⇒(y/e^(3x) ) = −ln ∣x∣ + C_1 x+C_2 ⇒y = −e^(3x) .ln ∣x∣ + (C_1 x+C_2 ).e^(3x)$
	$l e t y = e^{3 x} . q \to {\begin{cases} D y = e^{3 x} (q^{'} + 3 q) \\ D^{2} y = e^{3 x} (q^{″} + 6 q^{'} + 9 q) \end{cases}$ $w e o b t a i n$ $e^{3 x} (q^{″} + 6 q^{'} + 9 q) - 6 . e^{3 x} (q^{'} + 3 q) + 9 e^{3 x} q = \frac{e^{3 x}}{x^{2}}$ $w h i c h s i m p l i f i e s t o$ $q^{″} = \frac{1}{x^{2}} \to \frac{d (d q)}{d x} = \frac{1}{x^{2}}$ $\int \frac{d (d q)}{d x} = \int \frac{1}{x^{2}} \Rightarrow \frac{d q}{d x} = - \frac{1}{x} + C_{1}$ $\int d q = \int (- \frac{1}{x} + C_{1}) d x$ $\to q = - \ln ∣ x ∣ + C_{1} x + C_{2}$ $\Rightarrow \frac{y}{e^{3 x}} = - \ln ∣ x ∣ + C_{1} x + C_{2}$ $\Rightarrow y = - e^{3 x} . \ln ∣ x ∣ + (C_{1} x + C_{2}) . e^{3 x}$
	Answered by Bird last updated on 28/Sep/20

	$y^{″} - 6 y^{'} + 9 y = \frac{e^{3 x}}{x^{2}}$ $h \to r^{2} - 6 r + 9 = 0 \Rightarrow {(r - 3)}^{2} = 0 \Rightarrow$ $r = 3 \Rightarrow y_{h} = (a x + b) e^{3 x} = {a x e}^{3 x} + {b e}^{3 x}$ $= {a u}_{1} + {b u}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} {x e}^{3 x} e^{3 x} \\ (1 + 3 x) e^{3 x} 3 e^{3 x} \end{matrix} \|$ $= 3 x e^{6 x} - (1 + 3 x) e^{6 x} = - e^{6 x} \neq 0$ $W_{1} = \| \begin{matrix} o e^{3 x} \\ \frac{e^{3 x}}{x^{2}} 3 e^{3 x} \end{matrix} \| = - \frac{e^{6 x}}{x^{2}}$ $W_{2} = \| \begin{matrix} {x e}^{3 x} 0 \\ (1 + 3 x) e^{3 x} \frac{e^{3 x}}{x^{2}} \end{matrix} \| = \frac{e^{6 x}}{x}$ $v_{1} = \int \frac{w_{1}}{w} d x = \int - \frac{e^{6 x}}{x^{2} (- e^{6 x})}$ $= \int \frac{d x}{x^{2}} = - \frac{1}{x}$ $v_{2} = \int \frac{w_{2}}{w} d x = \int \frac{e^{6 x}}{- {x e}^{6 x}} d x$ $= - \int \frac{d x}{x} = - l n ∣ x ∣ \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2}$ $= {x e}^{3 x} (- \frac{1}{x}) - e^{3 x} l n ∣ x ∣$ $= - e^{3 x} - e^{3 x} l n ∣ x ∣ \Rightarrow t h e g e n e r a l$ $s o l u t i o n i s$ $y = {a x e}^{3 x} + {b e}^{3 x} - (1 + l n ∣ x ∣) e^{3 x}$ $= (a x + b - 1 - l n ∣ x ∣) e^{3 x}$
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