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Question Number 183533 by MikeH last updated on 26/Dec/22

Commented by MikeH last updated on 26/Dec/22

$please guys solve using the method undetermined$ $coefficients or variation of parameters .$
	Answered by qaz last updated on 27/Dec/22

	$y_{p} = \frac{1}{(D - 1) (D + 2)} {x e}^{x} = \frac{1}{3} (\frac{1}{D - 1} - \frac{1}{D + 2}) {x e}^{x}$ $= \frac{1}{3} e^{x} (\frac{1}{D} - \frac{1}{D + 3}) x$ $= \frac{1}{3} e^{x} (\frac{1}{2} x^{2} - \frac{1}{3 (1 + \frac{D}{3})} x)$ $= \frac{1}{6} x^{2} e^{x} - \frac{1}{9} e^{x} (1 - \frac{D}{3} + . . .) x$ $= \frac{1}{6} x^{2} e^{x} - \frac{1}{9} {x e}^{x} + \frac{1}{27} e^{x}$ $\Rightarrow y = C_{1} e^{x} + C_{2} e^{- 2 x} + \frac{1}{6} x^{2} e^{x} - \frac{1}{9} {x e}^{x} + \frac{1}{27} e^{x}$
	Answered by FelipeLz last updated on 26/Dec/22
	$y′′+y′−2y = 0 y = e^r → r^2 e^r +re^r −2e^r = 0 r^2 +r−2 = 0 → r = { (1),((−2)) :} y_h (x) = c_1 y_1 (x)+c_2 y_2 (x) = c_1 e^x +c_2 e^(−2x) y_p (x) = u(x)y_1 (x)+v(x)y_2 (x) y_p (x) = u(x)e^x +v(x)e^(−2x) { ((u′(x)e^x +v′(x)e^(−2x) = 0)),((u′(x)e^x +v′(x)(−2e^(−2x) ) = xe^x )) :} u′(x) = ( determinant ((( 0),( e^(−2x) )),((xe^x ),(−2e^(−2x) )))/ determinant ((e^x ,e^(−2x) ),(e^x ,(−2e^(−2x) )))) = ((−xe^(−x) )/(−2e^(−x) −e^(−x) )) = (x/3) u(x) = (1/3)∫xdx = (x^2 /6) v′(x) = ( determinant ((e^x ,0),(e^x ,(xe^x )))/ determinant ((e^x ,e^(−2x) ),(e^x ,(−2e^(−2x) )))) = ((xe^(2x) )/(−2e^(−x) −e^(−x) )) = −((xe^(3x) )/3) v(x) = −(1/3)∫xe^(3x) dx = −(1/3)[(1/3)xe^(3x) −(1/3)∫e^(3x) dx] = −(1/9)xe^(3x) +(1/(27))e^(3x) y_p (x) = (x^2 /6)e^x +(−(x/9)e^(3x) +(1/(27))e^(3x) )e^(−2x) y_p (x) = (x^2 /6)e^x −(x/9)e^x +(1/(27))e^x y(x) = y_h (x)+y_p (x) y(x) = ((x^2 /6)−(x/9)+(1/(27))+c_1 )e^x +c_2 e^(−2x) o$
	$y^{″} + y^{'} - 2 y = 0$ $y = e^{r} \to r^{2} e^{r} + {r e}^{r} - 2 e^{r} = 0$ $r^{2} + r - 2 = 0 \to r = {\begin{cases} 1 \\ - 2 \end{cases}$ $y_{h} (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x) = c_{1} e^{x} + c_{2} e^{- 2 x}$ $y_{p} (x) = u (x) y_{1} (x) + v (x) y_{2} (x)$ $y_{p} (x) = u (x) e^{x} + v (x) e^{- 2 x}$ ${\begin{cases} u^{'} (x) e^{x} + v^{'} (x) e^{- 2 x} = 0 \\ u^{'} (x) e^{x} + v^{'} (x) (- 2 e^{- 2 x}) = {x e}^{x} \end{cases}$ $u^{'} (x) = \frac{\| \begin{matrix} 0 & e^{- 2 x} \\ {x e}^{x} & - 2 e^{- 2 x} \end{matrix} \|}{\| \begin{matrix} e^{x} & e^{- 2 x} \\ e^{x} & - 2 e^{- 2 x} \end{matrix} \|} = \frac{- {x e}^{- x}}{- 2 e^{- x} - e^{- x}} = \frac{x}{3}$ $u (x) = \frac{1}{3} \int x d x = \frac{x^{2}}{6}$ $v^{'} (x) = \frac{\| \begin{matrix} e^{x} & 0 \\ e^{x} & {x e}^{x} \end{matrix} \|}{\| \begin{matrix} e^{x} & e^{- 2 x} \\ e^{x} & - 2 e^{- 2 x} \end{matrix} \|} = \frac{{x e}^{2 x}}{- 2 e^{- x} - e^{- x}} = - \frac{{x e}^{3 x}}{3}$ $v (x) = - \frac{1}{3} \int {x e}^{3 x} d x = - \frac{1}{3} [\frac{1}{3} {x e}^{3 x} - \frac{1}{3} \int e^{3 x} d x] = - \frac{1}{9} {x e}^{3 x} + \frac{1}{27} e^{3 x}$ $y_{p} (x) = \frac{x^{2}}{6} e^{x} + (- \frac{x}{9} e^{3 x} + \frac{1}{27} e^{3 x}) e^{- 2 x}$ $y_{p} (x) = \frac{x^{2}}{6} e^{x} - \frac{x}{9} e^{x} + \frac{1}{27} e^{x}$ $y (x) = y_{h} (x) + y_{p} (x)$ $y (x) = (\frac{x^{2}}{6} - \frac{x}{9} + \frac{1}{27} + c_{1}) e^{x} + c_{2} e^{- 2 x} o$
	Commented by MikeH last updated on 26/Dec/22

	$thank you so much$
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