solve (x−1)y^′ +(1+(√x))y =x e^(−2x)


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Question Number 57415 by Abdo msup. last updated on 03/Apr/19

$s o l v e (x - 1) y^{^{'}} + (1 + \sqrt{x}) y = x e^{- 2 x}$
Commented by maxmathsup by imad last updated on 12/Apr/19
$due to (√x)we must have x≥0 (ed) ⇔((√x)−1)((√x)+1)y^′ +((√( x))+1)y =xe^(−2x) ⇒ ((√x)−1)y^′ +y =(x/((√x)+1)) e^(−2x) (e) (he) ⇒((√x)−1)y^′ +y =0 ⇒((√x)−1)y^′ =−y ⇒(y^′ /y) =−(1/((√x)−1)) ⇒ ln∣y∣ =−∫ (dx/((√x)−1)) =_((√x)=t) −∫ ((2tdt)/(t−1)) =−2 ∫ ((t−1+1)/(t−1)) dt =−2t −2ln∣t−1∣ +c ⇒ y(x) =(K/((t−1)^2 )) e^(−2t) =(K/(((√x)−1)^2 )) e^(−2(√x)) let use mvc method we have y^′ (x) =K^′ ((√x)−1)^(−2) e^(−2(√x)) +K { −2 (1/(2(√x)))((√x)−1)^(−3) e^(−2(√x)) −2 (1/(2(√x))) ((√x)−1)^(−2) e^(−2(√x)) } =K^′ ((√x)−1)^(−2) e^(−2(√x)) −(K/(√x)){ (1/(((√x)−1)^3 )) +(1/(((√x)−1)^2 ))}e^(−2(√x)) and (e) ⇒ ((√x)−1)(K^′ /(((√x)−1)^2 ))−(K/(√x)){ (1/(((√x)−1)^2 )) +(1/((√x)−1))} +(K/(((√x)−1)^2 )) =(x/((√x) +1)) ⇒ (K^′ /((√x)−1)) −(K/(√x)){((√x)/(((√x)−1)^2 ))} +(K/(((√x)−1)^2 )) =(x/((√x) +1)) ⇒K^′ =((x((√x)−1))/((√x)+1)) ⇒ K(x) =∫ ((x((√x)−1))/((√x)+1)) dx +c but ∫ ((x((√x)−1))/((√x)+1)) dx =_((√x)=t) ∫ ((t^2 (t−1))/(t+1))(2t)dt =2 ∫ ((t^3 −t^2 )/(t+1)) dt =2 ∫ ((t^3 +1 −t^2 −1)/(t+1)) dt =2 ∫ (t^2 −t+1) −2 ∫ ((t^2 +1)/(t+1)) dt =(2/3)t^3 −t^2 +2t −2 ∫ ((t^2 −1 +2)/(t+1)) dt =(2/3)t^3 −t^2 +2t −2 ∫(t−1)dt −4ln∣t+1∣ =(2/3)t^3 −2t^2 +4t −4ln∣t+1∣ =(2/3)((√x))^3 −2x +4(√x) −4ln∣1+(√x)∣ ⇒ K(x) =((2x)/3)(√x)−2x+4(√x) −4ln(1+(√x))+c ⇒ y(x) =(e^(−2(√x)) /(((√x)−1)^2 )){ (2/3)x(√x)−2x +4(√x)−4ln(1+(√x)) +C} .$
$d u e t o \sqrt{x} w e m u s t h a v e x ⩾ 0 (e d) \Leftrightarrow (\sqrt{x} - 1) (\sqrt{x} + 1) y^{^{'}} + (\sqrt{x} + 1) y = {x e}^{- 2 x} \Rightarrow$ $(\sqrt{x} - 1) y^{^{'}} + y = \frac{x}{\sqrt{x} + 1} e^{- 2 x} (e)$ $(h e) \Rightarrow (\sqrt{x} - 1) y^{^{'}} + y = 0 \Rightarrow (\sqrt{x} - 1) y^{^{'}} = - y \Rightarrow \frac{y^{^{'}}}{y} = - \frac{1}{\sqrt{x} - 1} \Rightarrow$ $l n ∣ y ∣ = - \int \frac{d x}{\sqrt{x} - 1} =_{\sqrt{x} = t} - \int \frac{2 t d t}{t - 1} = - 2 \int \frac{t - 1 + 1}{t - 1} d t = - 2 t - 2 l n ∣ t - 1 ∣ + c \Rightarrow$ $y (x) = \frac{K}{{(t - 1)}^{2}} e^{- 2 t} = \frac{K}{{(\sqrt{x} - 1)}^{2}} e^{- 2 \sqrt{x}} l e t u s e m v c m e t h o d w e h a v e$ $y^{^{'}} (x) = K^{^{'}} {(\sqrt{x} - 1)}^{- 2} e^{- 2 \sqrt{x}} + K {- 2 \frac{1}{2 \sqrt{x}} {(\sqrt{x} - 1)}^{- 3} e^{- 2 \sqrt{x}} - 2 \frac{1}{2 \sqrt{x}} {(\sqrt{x} - 1)}^{- 2} e^{- 2 \sqrt{x}}}$ $= K^{^{'}} {(\sqrt{x} - 1)}^{- 2} e^{- 2 \sqrt{x}} - \frac{K}{\sqrt{x}} {\frac{1}{{(\sqrt{x} - 1)}^{3}} + \frac{1}{{(\sqrt{x} - 1)}^{2}}} e^{- 2 \sqrt{x}} a n d (e) \Rightarrow$ $(\sqrt{x} - 1) \frac{K^{^{'}}}{{(\sqrt{x} - 1)}^{2}} - \frac{K}{\sqrt{x}} {\frac{1}{{(\sqrt{x} - 1)}^{2}} + \frac{1}{\sqrt{x} - 1}} + \frac{K}{{(\sqrt{x} - 1)}^{2}} = \frac{x}{\sqrt{x} + 1} \Rightarrow$ $\frac{K^{^{'}}}{\sqrt{x} - 1} - \frac{K}{\sqrt{x}} {\frac{\sqrt{x}}{{(\sqrt{x} - 1)}^{2}}} + \frac{K}{{(\sqrt{x} - 1)}^{2}} = \frac{x}{\sqrt{x} + 1} \Rightarrow K^{^{'}} = \frac{x (\sqrt{x} - 1)}{\sqrt{x} + 1} \Rightarrow$ $K (x) = \int \frac{x (\sqrt{x} - 1)}{\sqrt{x} + 1} d x + c b u t$ $\int \frac{x (\sqrt{x} - 1)}{\sqrt{x} + 1} d x =_{\sqrt{x} = t} \int \frac{t^{2} (t - 1)}{t + 1} (2 t) d t = 2 \int \frac{t^{3} - t^{2}}{t + 1} d t$ $= 2 \int \frac{t^{3} + 1 - t^{2} - 1}{t + 1} d t = 2 \int (t^{2} - t + 1) - 2 \int \frac{t^{2} + 1}{t + 1} d t$ $= \frac{2}{3} t^{3} - t^{2} + 2 t - 2 \int \frac{t^{2} - 1 + 2}{t + 1} d t$ $= \frac{2}{3} t^{3} - t^{2} + 2 t - 2 \int (t - 1) d t - 4 l n ∣ t + 1 ∣$ $= \frac{2}{3} t^{3} - 2 t^{2} + 4 t - 4 l n ∣ t + 1 ∣ = \frac{2}{3} {(\sqrt{x})}^{3} - 2 x + 4 \sqrt{x} - 4 l n ∣ 1 + \sqrt{x} ∣ \Rightarrow$ $K (x) = \frac{2 x}{3} \sqrt{x} - 2 x + 4 \sqrt{x} - 4 l n (1 + \sqrt{x}) + c \Rightarrow$ $y (x) = \frac{e^{- 2 \sqrt{x}}}{{(\sqrt{x} - 1)}^{2}} {\frac{2}{3} x \sqrt{x} - 2 x + 4 \sqrt{x} - 4 l n (1 + \sqrt{x}) + C} .$
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