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Question Number 68873 by mathmax by abdo last updated on 16/Sep/19
solve (x^3 +1)y^′ +(2x+3)y =x cos(2x)
$${solve}\:\:\:\left({x}^{\mathrm{3}} +\mathrm{1}\right){y}^{'} \:+\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:={x}\:{cos}\left(\mathrm{2}{x}\right) \\ $$
Commented by mathmax by abdo last updated on 20/Sep/19
(he) → (x^3 +1)y^′ +(2x+3)y =0 ⇒(x^3 +1)y^′ =−(2x+3)y ⇒ (y^′ /y) =−((2x+3)/(x^3 +1)) ⇒ln∣y∣ =−∫ ((2x+3)/(x^3 +1))dx let decompose F(x)=((2x+3)/(x^3 +1)) ⇒F(x)=((2x+3)/((x+1)(x^2 −x+1))) =(a/(x+1)) +((bx+c)/(x^2 −x +1)) a =lim_(x→−1) (x+1)F(x)=(1/3) lim_(x→+∞) xF(x)=0 =a+b ⇒b=−(1/3) ⇒ F(x)=(1/(3(x+1))) +((−(1/3)x +c)/(x^2 −x +1)) F(o) =3 =(1/3) +c ⇒c =3−(1/3) =(8/3) ⇒F(x)=(1/(3(x+1)))−(1/3)((x−8)/(x^2 −x+1)) ⇒∫ F(x)dx =(1/3)ln∣x+1∣−(1/6) ∫((2x−1−15)/(x^2 −x+1))dx =(1/3)ln∣x+1∣−(1/6)ln(x^2 −x+1) +((15)/6) ∫ (dx/((x−(1/2))^(2 ) +(3/4))) ∫ (dx/((x−(1/2))^2 +(3/4))) =_(x−(1/2)=((√3)/2)t) (4/3) ∫ (1/(t^2 +1))((√3)/2)dt =(2/( (√3))) arctan(((2x−1)/( (√3)))) ⇒ln∣y∣ =−(1/3)ln∣x+1∣+(1/6)ln(x^2 −x+1)+((15)/(3(√3))) arctan(((2x−1)/( (√3)))) +c⇒ y =K ×(1/(∣x+1∣^(1/3) )) ×(x^2 −x+1)^(1/6) ×e^(((15)/(3(√3))) arctan(((2x−1)/( (√3))))) ...be continued...
$$\left({he}\right)\:\rightarrow\:\left({x}^{\mathrm{3}} +\mathrm{1}\right){y}^{'} \:+\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:=\mathrm{0}\:\Rightarrow\left({x}^{\mathrm{3}} \:+\mathrm{1}\right){y}^{'} =−\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=−\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{3}} \:+\mathrm{1}}\:\Rightarrow{ln}\mid{y}\mid\:=−\int\:\:\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{3}} \:+\mathrm{1}}{dx}\:\:{let}\:{decompose}\: \\ $$$${F}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{3}} \:+\mathrm{1}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\:=\frac{{a}}{{x}+\mathrm{1}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$${a}\:={lim}_{{x}\rightarrow−\mathrm{1}} \left({x}+\mathrm{1}\right){F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${lim}_{{x}\rightarrow+\infty} {xF}\left({x}\right)=\mathrm{0}\:={a}+{b}\:\Rightarrow{b}=−\frac{\mathrm{1}}{\mathrm{3}}\:\:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}\:+\frac{−\frac{\mathrm{1}}{\mathrm{3}}{x}\:+{c}}{{x}^{\mathrm{2}} −{x}\:+\mathrm{1}} \\ $$$${F}\left({o}\right)\:=\mathrm{3}\:=\frac{\mathrm{1}}{\mathrm{3}}\:+{c}\:\Rightarrow{c}\:=\mathrm{3}−\frac{\mathrm{1}}{\mathrm{3}}\:=\frac{\mathrm{8}}{\mathrm{3}}\:\Rightarrow{F}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{3}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{3}}\frac{{x}−\mathrm{8}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\Rightarrow\int\:{F}\left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{6}}\:\int\frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{15}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:+\frac{\mathrm{15}}{\mathrm{6}}\:\int\:\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}\:} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\int\:\:\frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}}\:=_{{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{t}} \:\:\:\:\frac{\mathrm{4}}{\mathrm{3}}\:\int\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} \:+\mathrm{1}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{dt}\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow{ln}\mid{y}\mid\:=−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{6}}{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\frac{\mathrm{15}}{\mathrm{3}\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\:+{c}\Rightarrow \\ $$$${y}\:={K}\:×\frac{\mathrm{1}}{\mid{x}+\mathrm{1}\mid^{\frac{\mathrm{1}}{\mathrm{3}}} }\:×\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} ×{e}^{\frac{\mathrm{15}}{\mathrm{3}\sqrt{\mathrm{3}}}\:{arctan}\left(\frac{\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)} \:\:…{be}\:{continued}… \\ $$

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