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

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

$${solve}\:{y}^{'} \sqrt{\mathrm{2}{x}−\mathrm{1}}\:+{y}\left({x}^{\mathrm{2}} +\mathrm{3}\right)\:={xsin}\left(\mathrm{2}{x}\right) \\ $$

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....

$$\left({he}\right)\:\rightarrow\sqrt{\mathrm{2}{x}−\mathrm{1}}{y}^{'} \:+\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}\:=\mathrm{0}\:\Rightarrow\sqrt{\mathrm{2}{x}−\mathrm{1}}{y}'=−\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=−\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:\Rightarrow{ln}\mid{y}\mid\:=−\int\:\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{dx}\:+{k}\:\Rightarrow \\ $$$${y}\left({x}\right)={C}\:{e}^{−\int\:\frac{{x}^{\mathrm{2}} +\mathrm{3}}{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{dx}} \:\:\:\:\:{changement}\:\sqrt{\mathrm{2}{x}−\mathrm{1}}={t}\:{give} \\ $$$$\mathrm{2}{x}−\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow\mathrm{2}{dx}\:=\mathrm{2}{tdt}\:\Rightarrow{dx}\:={tdt}\:{and} \\ $$$$\int\frac{{x}^{\mathrm{2}} \:+\mathrm{3}}{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{dx}\:=\int\:\:\frac{\left(\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{3}}{{t}}\:{tdt}\:=\int\left(\:\frac{\left({t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{3}\right){dt} \\ $$$$=\mathrm{3}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\:\left({t}^{\mathrm{4}} \:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}\right){dt}\:=\mathrm{3}{t}+\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\mathrm{1}}{\mathrm{5}}{t}^{\mathrm{5}} \:+\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+{t}\right\} \\ $$$$=\frac{\mathrm{13}}{\mathrm{4}}{t}\:+\frac{{t}^{\mathrm{5}} }{\mathrm{20}}\:+\frac{{t}^{\mathrm{3}} }{\mathrm{6}}\:=\frac{\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{5}} }{\mathrm{20}}\:+\frac{\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{3}} }{\mathrm{6}}\:+\frac{\mathrm{13}}{\mathrm{4}}\sqrt{\mathrm{2}{x}−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{20}}\:+\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{6}}\:+\frac{\mathrm{13}\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{4}}\:\Rightarrow \\ $$$${y}\left({x}\right)={C}\:{e}^{−\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{20}}−\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{6}}−\frac{\mathrm{13}\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{4}}} {let}\:{use}\:{mvc}\:{method} \\ $$$${letw}\left({x}\right)=\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{20}}+\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{6}}+\frac{\mathrm{13}\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{4}} \\ $$$$\Rightarrow{y}\left({x}\right)={C}\:{e}^{−{w}\left({x}\right)} \:\Rightarrow{y}^{'} \left({x}\right)\:={C}^{'} \:{e}^{−{w}\left({x}\right)} \:−{Cw}^{'} \left({x}\right)\:{e}^{−{w}\left({x}\right)} \\ $$$$=\left\{\:{C}^{\left(\mathrm{1}\right)} \:−{C}\:{w}^{'} \left({x}\right)\right\}{e}^{−{w}\left({x}\right)} \\ $$$${w}^{'} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{20}}×\mathrm{5}\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}}\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{4}} \:+\frac{\mathrm{1}}{\mathrm{6}}\mathrm{3}×\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}}\left(\sqrt{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{2}} \\ $$$$+\frac{\mathrm{13}}{\mathrm{4}}\:\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{2}{x}−\mathrm{1}}}\:=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \:+\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{4}\sqrt{\mathrm{2}{x}−\mathrm{1}}} \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{1}}{y}^{'} +\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){y}\:={xsin}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$$\sqrt{\mathrm{2}{x}−\mathrm{1}}\left\{\:{C}^{\left(\mathrm{1}\right)} \:−{C}\left(\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \:+\frac{\sqrt{\mathrm{2}{x}−\mathrm{1}}}{\mathrm{2}}\:+\frac{\mathrm{13}}{\mathrm{4}\sqrt{\mathrm{2}{x}−\mathrm{1}}}\right)\right\}{e}^{−{w}\left({x}\right)} \\ $$$$+\left({x}^{\mathrm{2}\:} +\mathrm{3}\right){C}\:{e}^{−{w}\left({x}\right)} \:={xsin}\left(\mathrm{2}{x}\right)\:\Rightarrow \\ $$$$\left\{\sqrt{\mathrm{2}{x}−\mathrm{1}}{C}^{\left(\mathrm{1}\right)} −\frac{{C}}{\mathrm{4}}\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{\mathrm{2}{x}−\mathrm{1}}−\frac{{C}}{\mathrm{2}}\left(\mathrm{2}{x}−\mathrm{1}\right)\:+\frac{\mathrm{13}}{\mathrm{4}}\:+\left({x}^{\mathrm{2}} \:+\mathrm{3}\right){C}\right\}{e}^{−{w}\left({x}\right)} \\ $$$$={xsin}\left(\mathrm{2}{x}\right).....{be}\:{continued}.... \\ $$$$ \\ $$$$ \\ $$

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