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Question Number 99600 by bemath last updated on 22/Jun/20

	Answered by mathmax by abdo last updated on 22/Jun/20
	$y^(′′) +3y^′ +2y =sin(e^x ) (he)→y^(′′) +3y^′ +2y =0 →r^2 +3r +2 =0 Δ =1 ⇒r_1 =((−3+1)/2) =−1 and r_2 =((−3−1)/2) =−2 ⇒y_h =a e^(−x) +be^(−2x) =au_1 +bu_2 W(u_1 ,u_2 ) = determinant (((e^(−x) e^(−2x) )),((−e^(−x) −2e^(−2x) )))=−2e^(−3x) +e^(−3x) =−e^(−3x) W_1 = determinant (((0 e^(−2x) )),((sin(e^x ) −2e^(−2x) )))=−e^(−2x) sin(e^x ) W_2 = determinant (((e^(−x) 0)),((−e^(−x) sin(e^x ))))=e^(−x) sin(e^x ) v_1 =∫ (w_1 /W)dx =∫ ((−e^(−2x) sin(e^x ))/(−e^(−3x) ))dx =∫ e^x sin(e^x )dx =−cos(e^x ) v_2 =∫ (W_2 /W)dx =∫ ((e^(−x) sin(e^x ))/(−e^(−3x) ))dx =−∫ e^(2x) sin(e^x )dx =_(e^x =t) =− ∫ t^2 ((sint)/t)dt =−∫t sint dt = −{ −tcost +∫ cost dt} =tcost −sint =e^x cos(e^x )−sin(e^x ) ⇒ y_p =u_1 v_1 +u_2 v_2 =−e^(−x) cos(e^x )+e^(−2x) (e^x cos(e^x )−sin(e^x )) =−e^(−x) cos(e^x )+e^(−x) cos(e^x )−e^(−2x) sin(e^x ) =−e^(−2x) sin(e^x ) the general solution is y =y_h +y_p =a e^(−x) +be^(−2x) −e^(−2x) sin(e^x )$
	$y^{^{″}} + {3 y}^{^{'}} + 2 y = \sin (e^{x})$ $(he) \to y^{^{″}} + {3 y}^{^{'}} + 2 y = 0 \to r^{2} + 3 r + 2 = 0$ $Δ = 1 \Rightarrow r_{1} = \frac{- 3 + 1}{2} = - 1 and r_{2} = \frac{- 3 - 1}{2} = - 2 \Rightarrow y_{h} = a e^{- x} + {be}^{- 2 x}$ $= {au}_{1} + {bu}_{2}$ $W (u_{1}, u_{2}) = \| \begin{matrix} e^{- x} e^{- 2 x} \\ - e^{- x} - {2 e}^{- 2 x} \end{matrix} \| = - {2 e}^{- 3 x} + e^{- 3 x} = - e^{- 3 x}$ $W_{1} = \| \begin{matrix} 0 e^{- 2 x} \\ \sin (e^{x}) - {2 e}^{- 2 x} \end{matrix} \| = - e^{- 2 x} \sin (e^{x})$ $W_{2} = \| \begin{matrix} e^{- x} 0 \\ - e^{- x} \sin (e^{x}) \end{matrix} \| = e^{- x} \sin (e^{x})$ $v_{1} = \int \frac{w_{1}}{W} dx = \int \frac{- e^{- 2 x} \sin (e^{x})}{- e^{- 3 x}} dx = \int e^{x} \sin (e^{x}) dx = - \cos (e^{x})$ $v_{2} = \int \frac{W_{2}}{W} dx = \int \frac{e^{- x} \sin (e^{x})}{- e^{- 3 x}} dx = - \int e^{2 x} \sin (e^{x}) dx =_{e^{x} = t}$ $= - \int t^{2} \frac{sint}{t} dt = - \int t sint dt = - {- tcost + \int cost dt}$ $= tcost - sint = e^{x} \cos (e^{x}) - \sin (e^{x}) \Rightarrow$ $y_{p} = u_{1} v_{1} + u_{2} v_{2} = - e^{- x} \cos (e^{x}) + e^{- 2 x} (e^{x} \cos (e^{x}) - \sin (e^{x}))$ $= - e^{- x} \cos (e^{x}) + e^{- x} \cos (e^{x}) - e^{- 2 x} \sin (e^{x}) = - e^{- 2 x} \sin (e^{x})$ $the general solution is y = y_{h} + y_{p} = a e^{- x} + {be}^{- 2 x} - e^{- 2 x} \sin (e^{x})$
	Commented by bemath last updated on 22/Jun/20

	$thank you sir$
	Commented by mathmax by abdo last updated on 22/Jun/20

	$you are welcome sir .$
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