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Question Number 157033 by amin96 last updated on 18/Oct/21

$${y}''=−{y} \\ $$

Commented by aliyn last updated on 19/Oct/21

y^(′′) +y = 0 ⇒ ( m−i) (m+i) = 0 ⇒ m=±i ∴ y = c_1 cos x +c_2 sin x □ M . T

$$\boldsymbol{{y}}^{''} \:+\boldsymbol{{y}}\:=\:\mathrm{0}\:\Rightarrow\:\left(\:\boldsymbol{{m}}−\boldsymbol{{i}}\right)\:\left(\boldsymbol{{m}}+\boldsymbol{{i}}\right)\:=\:\mathrm{0}\:\Rightarrow\:\boldsymbol{{m}}=\pm\boldsymbol{{i}} \\ $$$$ \\ $$$$\therefore\:\boldsymbol{{y}}\:=\:\boldsymbol{{c}}_{\mathrm{1}} \:\boldsymbol{{cos}}\:\boldsymbol{{x}}\:+\boldsymbol{{c}}_{\mathrm{2}} \:\boldsymbol{{sin}}\:\boldsymbol{{x}} \\ $$$$ \\ $$$$\square\:\boldsymbol{{M}}\:.\:\boldsymbol{{T}} \\ $$

Answered by puissant last updated on 18/Oct/21

y′′=−y → y′′+y=0 y=e^(rx) →y′=re^(rx) →y′′=r^2 e^(rx) y′′+y=e^(rx) (r^2 +1)=0 → r^2 +1=0 ⇒ r_1 =i=0+i ; r_2 =−i=0−i ⇒ y=(Acos(ix)+Bsin(ix))e^(0x) ⇒ y=(Acos(ix)+Bsin(ix))

$${y}''=−{y}\:\rightarrow\:{y}''+{y}=\mathrm{0}\: \\ $$$${y}={e}^{{rx}} \rightarrow{y}'={re}^{{rx}} \rightarrow{y}''={r}^{\mathrm{2}} {e}^{{rx}} \\ $$$${y}''+{y}={e}^{{rx}} \left({r}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0}\:\rightarrow\:{r}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\:{r}_{\mathrm{1}} ={i}=\mathrm{0}+{i}\:;\:\:{r}_{\mathrm{2}} =−{i}=\mathrm{0}−{i} \\ $$$$\Rightarrow\:{y}=\left({Acos}\left({ix}\right)+{Bsin}\left({ix}\right)\right){e}^{\mathrm{0}{x}} \\ $$$$\Rightarrow\:{y}=\left({Acos}\left({ix}\right)+{Bsin}\left({ix}\right)\right) \\ $$

Answered by amin96 last updated on 18/Oct/21

(d^2 y/dx^2 )=r^2 y=1 r^2 =i^2 ⇒ { ((r_1 =i)),((r_2 =−i)) :} ⇒ { ((Im{r}=1=a)),((Re{r}=0=b)) :} y=c_1 x^0 (cos(ax))+c_2 x^0 (sin(ax)) y=c_1 cos(x)+c_2 sin(x)

$$\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }=\boldsymbol{\mathrm{r}}^{\mathrm{2}} \:\:\:\:\boldsymbol{\mathrm{y}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{r}}^{\mathrm{2}} =\boldsymbol{\mathrm{i}}^{\mathrm{2}} \:\Rightarrow\begin{cases}{\boldsymbol{\mathrm{r}}_{\mathrm{1}} =\boldsymbol{\mathrm{i}}}\\{\boldsymbol{\mathrm{r}}_{\mathrm{2}} =−\boldsymbol{\mathrm{i}}}\end{cases}\:\:\:\Rightarrow\begin{cases}{{Im}\left\{{r}\right\}=\mathrm{1}=\boldsymbol{\mathrm{a}}}\\{{Re}\left\{{r}\right\}=\mathrm{0}=\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$${y}=\boldsymbol{\mathrm{c}}_{\mathrm{1}} {x}^{\mathrm{0}} \left(\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{ax}}\right)\right)+\boldsymbol{\mathrm{c}}_{\mathrm{2}} {x}^{\mathrm{0}} \left(\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{ax}}\right)\right) \\ $$$$\boldsymbol{\mathrm{y}}=\boldsymbol{\mathrm{c}}_{\mathrm{1}} \boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{x}}\right)+\boldsymbol{\mathrm{c}}_{\mathrm{2}} \boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{x}}\right) \\ $$

Answered by mr W last updated on 18/Oct/21

say y′=(dy/dx)=u y′′=(du/dx)=u(du/dy) u(du/dy)=−y ∫udu=−∫ydy u^2 =C^2 −y^2 u=(dy/dx)=±(√(C^2 −y^2 )) ∫(dy/( (√(C^2 −y^2 ))))=±∫dx sin^(−1) (y/C)=±(x+C_2 ) ⇒y=C_1 sin (x+C_2 )

$${say}\:{y}'=\frac{{dy}}{{dx}}={u} \\ $$$${y}''=\frac{{du}}{{dx}}={u}\frac{{du}}{{dy}} \\ $$$${u}\frac{{du}}{{dy}}=−{y} \\ $$$$\int{udu}=−\int{ydy} \\ $$$${u}^{\mathrm{2}} ={C}^{\mathrm{2}} −{y}^{\mathrm{2}} \\ $$$${u}=\frac{{dy}}{{dx}}=\pm\sqrt{{C}^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$\int\frac{{dy}}{\:\sqrt{{C}^{\mathrm{2}} −{y}^{\mathrm{2}} }}=\pm\int{dx} \\ $$$$\mathrm{sin}^{−\mathrm{1}} \frac{{y}}{{C}}=\pm\left({x}+{C}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{y}={C}_{\mathrm{1}} \:\mathrm{sin}\:\left({x}+{C}_{\mathrm{2}} \right) \\ $$

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