Menu Close

solve-the-de-x-2-1-y-xy-x-2-x-

Tinku Tara 0 Comments
Pin




Question Number 65402 by mathmax by abdo last updated on 29/Jul/19
solve the (de) (√(x^2 +1))y^(′′) −xy^′ =x^2 −x
$${solve}\:{the}\:\left({de}\right)\:\:\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{y}^{''} \:\:\:−{xy}^{'} \:={x}^{\mathrm{2}} −{x} \\ $$
Commented by mathmax by abdo last updated on 30/Jul/19
let y^′ =z (e) ⇒(√(x^2 +1))z^′ −xz =x^2 −x (he) ⇒(√(x^2 +1))z^′ −xz =0 ⇒(√(x^2 +1))z^′ =xz ⇒(z^′ /z) =(x/( (√(x^2 +1)))) ⇒ ln∣z∣ =(√(x^2 +1 )) +c ⇒z =k e^(√(1+x^2 )) mvc method give z^′ =k^′ e^(√(1+x^2 )) +k (x/( (√(1+x^2 ))))e^(√(1+x^2 )) (e) ⇒(√(1+x^2 ))( k^′ +((kx)/( (√(1+x^2 )))))e^(√(1+x^2 )) −xke^(√(1+x^2 )) =x^2 −x ⇒ (√(1+x^2 ))k^′ e^(√(1+x^2 )) +kxe^(√(1+x^2 )) −xk e^(√(1+x^2 )) =x^2 −x ⇒ k^′ =((x^2 −x)/( (√(1+x^2 )))) e^(−(√(1+x^2 ))) ⇒k(x) =∫ ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 ))) dx +c ⇒ z(x) =( ∫ ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 ))) dx +c)e^(√(1+x^2 )) =e^(√(1+x^2 )) ∫ ((x^2 −x)/( (√(1+x^2 ))))e^(−(√(1+x^2 ))) dx +c e^(√(1+x^2 )) y^′ =z(x) =⇒y(x) =∫^x z(t)dt +λ =∫^x {e^(√(1+t^2 )) ∫^t ((u^2 −u)/( (√(1+u^2 ))))e^(−(√(1+u^2 ))) du +c e^(√(1+t^2 )) }dt +λ
$${let}\:{y}^{'} ={z}\:\:\:\:\left({e}\right)\:\Rightarrow\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{z}^{'} −{xz}\:={x}^{\mathrm{2}} −{x} \\ $$$$\left({he}\right)\:\Rightarrow\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{z}^{'} −{xz}\:=\mathrm{0}\:\Rightarrow\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}{z}^{'} ={xz}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\:\Rightarrow \\ $$$${ln}\mid{z}\mid\:=\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}\:}\:+{c}\:\Rightarrow{z}\:={k}\:{e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${mvc}\:{method}\:{give}\:\:{z}^{'} \:={k}^{'} {e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:+{k}\:\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\left({e}\right)\:\Rightarrow\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\left(\:{k}^{'} \:+\frac{{kx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right){e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} −{xke}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:={x}^{\mathrm{2}} −{x}\:\Rightarrow \\ $$$$\sqrt{\mathrm{1}+{x}^{\mathrm{2}} \:}{k}^{'} \:{e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:+{kxe}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} −{xk}\:{e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} ={x}^{\mathrm{2}} −{x}\:\Rightarrow \\ $$$${k}^{'} \:=\frac{{x}^{\mathrm{2}} −{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{e}^{−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:\:\Rightarrow{k}\left({x}\right)\:=\int\:\:\:\frac{{x}^{\mathrm{2}} −{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{e}^{−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:{dx}\:+{c}\:\Rightarrow \\ $$$${z}\left({x}\right)\:=\left(\:\int\:\frac{{x}^{\mathrm{2}} −{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{e}^{−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} {dx}\:+{c}\right){e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$={e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \:\int\:\:\:\frac{{x}^{\mathrm{2}} −{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{e}^{−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} {dx}\:\:+{c}\:{e}^{\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$${y}^{'} \:={z}\left({x}\right)\:=\Rightarrow{y}\left({x}\right)\:=\int^{{x}} {z}\left({t}\right){dt}\:+\lambda \\ $$$$=\int^{{x}} \:\left\{{e}^{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \:\int^{{t}} \:\frac{{u}^{\mathrm{2}} −{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}{e}^{−\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} {du}\:\:+{c}\:{e}^{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }} \right\}{dt}\:+\lambda\: \\ $$

Pin

Leave a Reply

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