solve y^′ (√(2x−1)) +y(x^2 +3) =xsin(2x)


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Question Number 63824 by mathmax by abdo last updated on 10/Jul/19

$s o l v e y^{^{'}} \sqrt{2 x - 1} + y (x^{2} + 3) = x s i n (2 x)$
Commented by mathmax by abdo last updated on 12/Jul/19
$(he) →(√(2x−1))y^′ +(x^2 +3)y =0 ⇒(√(2x−1))y′=−(x^2 +3)y ⇒ (y^′ /y) =−((x^2 +3)/(√(2x−1))) ⇒ln∣y∣ =−∫ ((x^2 +3)/(√(2x−1)))dx +k ⇒ y(x)=C e^(−∫ ((x^2 +3)/(√(2x−1)))dx) changement (√(2x−1))=t give 2x−1 =t^2 ⇒2dx =2tdt ⇒dx =tdt and ∫((x^2 +3)/(√(2x−1)))dx =∫ (((((t^2 +1)/2))^2 +3)/t) tdt =∫( (((t^2 +1)^2 )/4)+3)dt =3t +(1/4) ∫ (t^4 +2t^2 +1)dt =3t+(1/4){(1/5)t^5 +(2/3)t^3 +t} =((13)/4)t +(t^5 /(20)) +(t^3 /6) =((((√(2x−1)))^5 )/(20)) +((((√(2x−1)))^3 )/6) +((13)/4)(√(2x−1)) =(((2x−1)^2 (√(2x−1)))/(20)) +(((2x−1)(√(2x−1)))/6) +((13(√(2x−1)))/4) ⇒ y(x)=C e^(−(((2x−1)^2 (√(2x−1)))/(20))−(((2x−1)(√(2x−1)))/6)−((13(√(2x−1)))/4)) let use mvc method letw(x)=(((2x−1)^2 (√(2x−1)))/(20))+(((2x−1)(√(2x−1)))/6)+((13(√(2x−1)))/4) ⇒y(x)=C e^(−w(x)) ⇒y^′ (x) =C^′ e^(−w(x)) −Cw^′ (x) e^(−w(x)) ={ C^((1)) −C w^′ (x)}e^(−w(x)) w^′ (x) =(1/(20))×5(2/(2(√(2x−1))))((√(2x−1)))^4 +(1/6)3×(2/(2(√(2x−1))))((√(2x−1)))^2 +((13)/4) (2/(2(√(2x−1)))) =(1/4)(2x−1)^2 +((√(2x−1))/2) +((13)/(4(√(2x−1)))) (√(2x−1))y^′ +(x^2 +3)y =xsin(2x) ⇒ (√(2x−1)){ C^((1)) −C((1/4)(2x−1)^2 +((√(2x−1))/2) +((13)/(4(√(2x−1)))))}e^(−w(x)) +(x^(2 ) +3)C e^(−w(x)) =xsin(2x) ⇒ {(√(2x−1))C^((1)) −(C/4)(2x−1)^2 (√(2x−1))−(C/2)(2x−1) +((13)/4) +(x^2 +3)C}e^(−w(x)) =xsin(2x).....be continued....$
$(h e) \to \sqrt{2 x - 1} y^{^{'}} + (x^{2} + 3) y = 0 \Rightarrow \sqrt{2 x - 1} y^{'} = - (x^{2} + 3) y \Rightarrow$ $\frac{y^{^{'}}}{y} = - \frac{x^{2} + 3}{\sqrt{2 x - 1}} \Rightarrow l n ∣ y ∣ = - \int \frac{x^{2} + 3}{\sqrt{2 x - 1}} d x + k \Rightarrow$ $y (x) = C e^{- \int \frac{x^{2} + 3}{\sqrt{2 x - 1}} d x} c h a n g e m e n t \sqrt{2 x - 1} = t g i v e$ $2 x - 1 = t^{2} \Rightarrow 2 d x = 2 t d t \Rightarrow d x = t d t a n d$ $\int \frac{x^{2} + 3}{\sqrt{2 x - 1}} d x = \int \frac{{(\frac{t^{2} + 1}{2})}^{2} + 3}{t} t d t = \int (\frac{{(t^{2} + 1)}^{2}}{4} + 3) d t$ $= 3 t + \frac{1}{4} \int (t^{4} + 2 t^{2} + 1) d t = 3 t + \frac{1}{4} {\frac{1}{5} t^{5} + \frac{2}{3} t^{3} + t}$ $= \frac{13}{4} t + \frac{t^{5}}{20} + \frac{t^{3}}{6} = \frac{{(\sqrt{2 x - 1})}^{5}}{20} + \frac{{(\sqrt{2 x - 1})}^{3}}{6} + \frac{13}{4} \sqrt{2 x - 1}$ $= \frac{{(2 x - 1)}^{2} \sqrt{2 x - 1}}{20} + \frac{(2 x - 1) \sqrt{2 x - 1}}{6} + \frac{13 \sqrt{2 x - 1}}{4} \Rightarrow$ $y (x) = C e^{- \frac{{(2 x - 1)}^{2} \sqrt{2 x - 1}}{20} - \frac{(2 x - 1) \sqrt{2 x - 1}}{6} - \frac{13 \sqrt{2 x - 1}}{4}} l e t u s e m v c m e t h o d$ $l e t w (x) = \frac{{(2 x - 1)}^{2} \sqrt{2 x - 1}}{20} + \frac{(2 x - 1) \sqrt{2 x - 1}}{6} + \frac{13 \sqrt{2 x - 1}}{4}$ $\Rightarrow y (x) = C e^{- w (x)} \Rightarrow y^{^{'}} (x) = C^{^{'}} e^{- w (x)} - {C w}^{^{'}} (x) e^{- w (x)}$ $= {C^{(1)} - C w^{^{'}} (x)} e^{- w (x)}$ $w^{^{'}} (x) = \frac{1}{20} \times 5 \frac{2}{2 \sqrt{2 x - 1}} {(\sqrt{2 x - 1})}^{4} + \frac{1}{6} 3 \times \frac{2}{2 \sqrt{2 x - 1}} {(\sqrt{2 x - 1})}^{2}$ $+ \frac{13}{4} \frac{2}{2 \sqrt{2 x - 1}} = \frac{1}{4} {(2 x - 1)}^{2} + \frac{\sqrt{2 x - 1}}{2} + \frac{13}{4 \sqrt{2 x - 1}}$ $\sqrt{2 x - 1} y^{^{'}} + (x^{2} + 3) y = x s i n (2 x) \Rightarrow$ $\sqrt{2 x - 1} {C^{(1)} - C (\frac{1}{4} {(2 x - 1)}^{2} + \frac{\sqrt{2 x - 1}}{2} + \frac{13}{4 \sqrt{2 x - 1}})} e^{- w (x)}$ $+ (x^{2} + 3) C e^{- w (x)} = x s i n (2 x) \Rightarrow$ ${\sqrt{2 x - 1} C^{(1)} - \frac{C}{4} {(2 x - 1)}^{2} \sqrt{2 x - 1} - \frac{C}{2} (2 x - 1) + \frac{13}{4} + (x^{2} + 3) C} e^{- w (x)}$ $= x s i n (2 x) . . . . . b e c o n t i n u e d . . . .$
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