solve-x-2-y-xy-y-0-on-0-put-x-e-t-

Question Number 65044 by mathmax by abdo last updated on 24/Jul/19

$$\left.{solve}\:{x}^{\mathrm{2}} {y}^{''} \:+{xy}^{'} \:+{y}\:=\mathrm{0}\:\:{on}\:\right]\mathrm{0},+\infty\left[\:\:\:\left({put}\:{x}\:={e}^{{t}} \right)\right. \\ $$

Commented by mathmax by abdo last updated on 08/Aug/19

let x =e^t ⇒y(x)=y(e^t )=z(t)⇒y(x)=z(lnx) ⇒ y^′ =(1/x)z^′ (lnx) and y^(′′) =−(1/x^2 )z^′ (lnx)+(1/x^2 )z^(′′) (lnx) (e) ⇒x^2 (−(1/x^2 )z^′ (lnx)+(1/x^2 )z^(′′) (lnx))+z^′ (lnx) +z(lnx)=0 ⇒ ⇒z^(′′) (t)+z(t) =0 →ce→r^2 +1 =0 ⇒r=+^− i ⇒ z =α e^(it) +β e^(−it) ⇒z(t) =acost +bsint ⇒ y(x)=a cos(lnx)+bsin(lnx).

$${let}\:{x}\:={e}^{{t}} \:\Rightarrow{y}\left({x}\right)={y}\left({e}^{{t}} \right)={z}\left({t}\right)\Rightarrow{y}\left({x}\right)={z}\left({lnx}\right)\:\Rightarrow \\ $$$${y}^{'} =\frac{\mathrm{1}}{{x}}{z}^{'} \left({lnx}\right)\:{and}\:{y}^{''} =−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{z}^{'} \left({lnx}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{z}^{''} \left({lnx}\right) \\ $$$$\left({e}\right)\:\Rightarrow{x}^{\mathrm{2}} \left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{z}^{'} \left({lnx}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{z}^{''} \left({lnx}\right)\right)+{z}^{'} \left({lnx}\right)\:+{z}\left({lnx}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow{z}^{''} \left({t}\right)+{z}\left({t}\right)\:=\mathrm{0}\:\:\:\:\rightarrow{ce}\rightarrow{r}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{r}=\overset{−} {+}{i}\:\Rightarrow \\ $$$${z}\:=\alpha\:{e}^{{it}} \:+\beta\:{e}^{−{it}} \:\Rightarrow{z}\left({t}\right)\:={acost}\:+{bsint}\:\Rightarrow \\ $$$${y}\left({x}\right)={a}\:{cos}\left({lnx}\right)+{bsin}\left({lnx}\right). \\ $$

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