Question Number 76842 by jagoll last updated on 31/Dec/19
![what is solution y^(′′ ) + y = 0 .](https://www.tinkutara.com/question/Q76842.png)
$$ \\ $$$${what}\:{is}\:{solution}\:{y}^{''\:} +\:\:{y}\:=\:\mathrm{0}\:. \\ $$
Commented by mathmax by abdo last updated on 31/Dec/19
![the caraceristc equation is r^2 +1=0 ⇒r=i or r=−i ⇒ y(x) =αe^(ix) +β e^(−ix) ⇒y(x)=acosx+ bsinx](https://www.tinkutara.com/question/Q76873.png)
$${the}\:{caraceristc}\:{equation}\:{is}\:{r}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:\Rightarrow{r}={i}\:{or}\:{r}=−{i}\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\alpha{e}^{{ix}} \:+\beta\:{e}^{−{ix}} \:\:\Rightarrow{y}\left({x}\right)={acosx}+\:{bsinx} \\ $$
Answered by mr W last updated on 31/Dec/19
![let y′=(dy/dx)=u y′′=((d(y′))/dx)=(du/dx)=(du/dy)×(dy/dx)=(du/dy)×u ⇒(du/dy)×u+y=0 ⇒udu=−ydy ⇒∫udu=−∫ydy ⇒(u^2 /2)=−(y^2 /2)+(c_1 ^2 /2) ⇒u=±(√(c_1 ^2 −y^2 )) ⇒(dy/dx)=±(√(c_1 ^2 −y^2 )) ⇒(dy/( (√(c_1 ^2 −y^2 ))))=±dx ⇒∫(dy/( (√(c_1 ^2 −y^2 ))))=±∫dx ⇒sin^(−1) (y/c_1 )=±x+c_2 ⇒y=c_1 sin (±x+c_2 ) ⇒y=c_1 sin (x+c_2 )](https://www.tinkutara.com/question/Q76848.png)
$${let}\:{y}'=\frac{{dy}}{{dx}}={u} \\ $$$${y}''=\frac{{d}\left({y}'\right)}{{dx}}=\frac{{du}}{{dx}}=\frac{{du}}{{dy}}×\frac{{dy}}{{dx}}=\frac{{du}}{{dy}}×{u} \\ $$$$\Rightarrow\frac{{du}}{{dy}}×{u}+{y}=\mathrm{0} \\ $$$$\Rightarrow{udu}=−{ydy} \\ $$$$\Rightarrow\int{udu}=−\int{ydy} \\ $$$$\Rightarrow\frac{{u}^{\mathrm{2}} }{\mathrm{2}}=−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+\frac{{c}_{\mathrm{1}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{u}=\pm\sqrt{{c}_{\mathrm{1}} ^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\pm\sqrt{{c}_{\mathrm{1}} ^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{dy}}{\:\sqrt{{c}_{\mathrm{1}} ^{\mathrm{2}} −{y}^{\mathrm{2}} }}=\pm{dx} \\ $$$$\Rightarrow\int\frac{{dy}}{\:\sqrt{{c}_{\mathrm{1}} ^{\mathrm{2}} −{y}^{\mathrm{2}} }}=\pm\int{dx} \\ $$$$\Rightarrow\mathrm{sin}^{−\mathrm{1}} \frac{{y}}{{c}_{\mathrm{1}} }=\pm{x}+{c}_{\mathrm{2}} \\ $$$$\Rightarrow{y}={c}_{\mathrm{1}} \mathrm{sin}\:\left(\pm{x}+{c}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{y}={c}_{\mathrm{1}} \mathrm{sin}\:\left({x}+{c}_{\mathrm{2}} \right) \\ $$
Commented by jagoll last updated on 31/Dec/19
![thanks you sir](https://www.tinkutara.com/question/Q76850.png)
$${thanks}\:{you}\:{sir} \\ $$
Commented by jagoll last updated on 31/Dec/19
![sir why not c_(1 ) as constan first integrate ? but (1/2)c_(1 ) ^( 2 )](https://www.tinkutara.com/question/Q76851.png)
$${sir}\:{why}\:{not}\:{c}_{\mathrm{1}\:} {as}\:{constan}\:{first}\:{integrate}\:?\: \\ $$$${but}\:\frac{\mathrm{1}}{\mathrm{2}}{c}_{\mathrm{1}\:} ^{\:\mathrm{2}\:} \\ $$
Commented by mr W last updated on 31/Dec/19
![it makes no real difference, but looks better in later operations.](https://www.tinkutara.com/question/Q76852.png)
$${it}\:{makes}\:{no}\:{real}\:{difference},\:{but} \\ $$$${looks}\:{better}\:{in}\:{later}\:{operations}. \\ $$
Commented by jagoll last updated on 31/Dec/19
![oo ok sir thanks you](https://www.tinkutara.com/question/Q76854.png)
$${oo}\:{ok}\:{sir}\:{thanks}\:{you} \\ $$
Answered by Rio Michael last updated on 18/Jan/20
![i could also solve this by (d^2 y/(dx^2 )) + y = 0 the auxiliary equation is r^2 + 1 = 0 ⇒ r= i or r = −i or r = 0 + i and r = 0−i the general solution is of the firm y = e^(px) (Acosqx +B sinqx) where p =0 and q=1 y = Acos x + Bsinx](https://www.tinkutara.com/question/Q78545.png)
$${i}\:{could}\:{also}\:{solve}\:{this}\:{by}\: \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} \:}\:+\:{y}\:=\:\mathrm{0} \\ $$$${the}\:{auxiliary}\:{equation}\:{is} \\ $$$${r}^{\mathrm{2}} \:+\:\mathrm{1}\:=\:\mathrm{0}\:\:\Rightarrow\:{r}=\:{i}\:{or}\:{r}\:=\:−{i} \\ $$$${or}\:\:{r}\:=\:\mathrm{0}\:+\:{i}\:{and}\:{r}\:=\:\mathrm{0}−{i}\: \\ $$$${the}\:{general}\:{solution} \\ $$$${is}\:{of}\:{the}\:{firm}\:\:{y}\:=\:{e}^{{px}} \left({Acosqx}\:+{B}\:{sinqx}\right) \\ $$$${where}\:{p}\:=\mathrm{0}\:{and}\:{q}=\mathrm{1} \\ $$$${y}\:=\:{Acos}\:{x}\:+\:{Bsinx} \\ $$