Menu Close

sove-the-de-x-2-y-2x-3-y-sin-x-2-with-y-1-2-and-y-1-1-




Question Number 63894 by mathmax by abdo last updated on 10/Jul/19
sove the (de) x^2 y^′  −(2x+3)y =sin(x^2 )  with y(1)=2 and  y^′ (1)=1 .
$${sove}\:{the}\:\left({de}\right)\:{x}^{\mathrm{2}} {y}^{'} \:−\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:={sin}\left({x}^{\mathrm{2}} \right)\:\:{with}\:{y}\left(\mathrm{1}\right)=\mathrm{2}\:{and} \\ $$$${y}^{'} \left(\mathrm{1}\right)=\mathrm{1}\:. \\ $$
Commented by mathmax by abdo last updated on 11/Jul/19
(he) →x^2 y^′ −(2x+3)y=0 ⇒x^2 y^′  =(2x+3)y ⇒  (y^′ /y) =((2x+3)/x^2 ) =(2/x) +(3/x^2 ) ⇒ln∣y∣=2ln∣x∣−(3/x)+c ⇒  y =K x^2  e^(−(3/x))    let use mvc method   y^′  =K^′ x^2  e^(−(3/x))  +K{ 2x e^(−(3/x))  +x^2 (3/x^2 ) e^(−(3/x)) }  =K^′  x^2  e^(−(3/(x )))  +K{2x+3}e^(−(3/x))   (e) ⇒K^′ x^4  e^(−(3/x))   +Kx^2 (2x+3)e^(−(3/x))  −(2x+3)Kx^2  e^(−(3/x))  =sin(x^2 ) ⇒  K^′  x^4  e^(−(3/x))  =sin(x^2 ) ⇒K^′  =((sin(x^2 )e^(3/x) )/x^4 ) ⇒  K(x) =∫_1 ^x    ((sin(t^2 )e^(3/t) )/t^4 )dt  +λ  λ=K(1) we have y(1)=K(1)e^(−3)  ⇒K(1)=e^3 y(1) ⇒  K(x) =∫_1 ^x   ((sin(t^2 )e^(3/t) )/t^4 )dt  +2e^3  ⇒  y(x) =x^2  e^(−(3/x)) { ∫_1 ^x   ((e^(3/t)  sin(t^2 ))/t^4 )dt +2e^3 }....
$$\left({he}\right)\:\rightarrow{x}^{\mathrm{2}} {y}^{'} −\left(\mathrm{2}{x}+\mathrm{3}\right){y}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} {y}^{'} \:=\left(\mathrm{2}{x}+\mathrm{3}\right){y}\:\Rightarrow \\ $$$$\frac{{y}^{'} }{{y}}\:=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} }\:=\frac{\mathrm{2}}{{x}}\:+\frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:\Rightarrow{ln}\mid{y}\mid=\mathrm{2}{ln}\mid{x}\mid−\frac{\mathrm{3}}{{x}}+{c}\:\Rightarrow \\ $$$${y}\:={K}\:{x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:\:\:{let}\:{use}\:{mvc}\:{method}\: \\ $$$${y}^{'} \:={K}^{'} {x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:+{K}\left\{\:\mathrm{2}{x}\:{e}^{−\frac{\mathrm{3}}{{x}}} \:+{x}^{\mathrm{2}} \frac{\mathrm{3}}{{x}^{\mathrm{2}} }\:{e}^{−\frac{\mathrm{3}}{{x}}} \right\} \\ $$$$={K}^{'} \:{x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}\:}} \:+{K}\left\{\mathrm{2}{x}+\mathrm{3}\right\}{e}^{−\frac{\mathrm{3}}{{x}}} \\ $$$$\left({e}\right)\:\Rightarrow{K}^{'} {x}^{\mathrm{4}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:\:+{Kx}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{3}\right){e}^{−\frac{\mathrm{3}}{{x}}} \:−\left(\mathrm{2}{x}+\mathrm{3}\right){Kx}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:={sin}\left({x}^{\mathrm{2}} \right)\:\Rightarrow \\ $$$${K}^{'} \:{x}^{\mathrm{4}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \:={sin}\left({x}^{\mathrm{2}} \right)\:\Rightarrow{K}^{'} \:=\frac{{sin}\left({x}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{x}}} }{{x}^{\mathrm{4}} }\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int_{\mathrm{1}} ^{{x}} \:\:\:\frac{{sin}\left({t}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{t}}} }{{t}^{\mathrm{4}} }{dt}\:\:+\lambda \\ $$$$\lambda={K}\left(\mathrm{1}\right)\:{we}\:{have}\:{y}\left(\mathrm{1}\right)={K}\left(\mathrm{1}\right){e}^{−\mathrm{3}} \:\Rightarrow{K}\left(\mathrm{1}\right)={e}^{\mathrm{3}} {y}\left(\mathrm{1}\right)\:\Rightarrow \\ $$$${K}\left({x}\right)\:=\int_{\mathrm{1}} ^{{x}} \:\:\frac{{sin}\left({t}^{\mathrm{2}} \right){e}^{\frac{\mathrm{3}}{{t}}} }{{t}^{\mathrm{4}} }{dt}\:\:+\mathrm{2}{e}^{\mathrm{3}} \:\Rightarrow \\ $$$${y}\left({x}\right)\:={x}^{\mathrm{2}} \:{e}^{−\frac{\mathrm{3}}{{x}}} \left\{\:\int_{\mathrm{1}} ^{{x}} \:\:\frac{{e}^{\frac{\mathrm{3}}{{t}}} \:{sin}\left({t}^{\mathrm{2}} \right)}{{t}^{\mathrm{4}} }{dt}\:+\mathrm{2}{e}^{\mathrm{3}} \right\}…. \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *