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Question Number 76842 by jagoll last updated on 31/Dec/19

$$$$$$whatissolution{y}^{{}^{″}}+y=0.$$

Commented by mathmax by abdo last updated on 31/Dec/19

$$thecaraceristcequationis{r}^2+1=0\Rightarrow r=iorr=-i\Rightarrow$$$$y\left(x\right)=\alpha e^{ix}+\beta e^{-ix}\Rightarrow y\left(x\right)=acosx+bsinx$$

Answered by mr W last updated on 31/Dec/19

$$let{y}^{\prime }=\frac{dy}{dx}=u$$$$y^{″}=\frac{d\left(y^{\prime }\right)}{dx}=\frac{du}{dx}=\frac{du}{dy}\times \frac{dy}{dx}=\frac{du}{dy}\times u$$$$\Rightarrow \frac{du}{dy}\times u+y=0$$$$\Rightarrow udu=-ydy$$$$\Rightarrow \int udu=-\int ydy$$$$\Rightarrow \frac{u^2}{2}=-\frac{y^2}{2}+\frac{c_1^2}{2}$$$$\Rightarrow u=\pm \sqrt{c_1^2-y^2}$$$$\Rightarrow \frac{dy}{dx}=\pm \sqrt{c_1^2-y^2}$$$$\Rightarrow \frac{dy}{\sqrt{c_1^2-y^2}}=\pm dx$$$$\Rightarrow \int \frac{dy}{\sqrt{c_1^2-y^2}}=\pm \int dx$$$$\Rightarrow {\sin }^{-1}\frac{y}{c_1}=\pm x+c_2$$$$\Rightarrow y=c_1\sin (\pm x+c_2)$$$$\Rightarrow y=c_1\sin (x+c_2)$$

Commented by jagoll last updated on 31/Dec/19

$$thanksyousir$$

Commented by jagoll last updated on 31/Dec/19

$$sirwhynot{c}_{1}asconstanfirstintegrate?$$$$but\frac{1}{2}{c}_1^2$$

Commented by mr W last updated on 31/Dec/19

$$itmakesnorealdifference,but$$$$looksbetterinlateroperations.$$

Commented by jagoll last updated on 31/Dec/19

$$oooksirthanksyou$$

Answered by Rio Michael last updated on 18/Jan/20

$$icouldalsosolvethisby$$$$\frac{d^{2}y}{{dx}^2}+y=0$$$$theauxiliaryequationis$$$$r^2+1=0\Rightarrow r=iorr=-i$$$$orr=0+iandr=0-i$$$$thegeneralsolution$$$$isofthefirmy=e^{px}(Acosqx+Bsinqx)$$$$wherep=0andq=1$$$$y=Acosx+Bsinx$$

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