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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

$${\mathit{y}}^{{}^{″}}+\mathit{y}=0\Rightarrow (\mathit{m}-\mathit{i})(\mathit{m}+\mathit{i})=0\Rightarrow \mathit{m}=\pm \mathit{i}$$$$$$$$\therefore \mathit{y}={\mathit{c}}_1\mathit{c}\mathit{o}\mathit{s}\mathit{x}+{\mathit{c}}_2\mathit{s}\mathit{i}\mathit{n}\mathit{x}$$$$$$$$◻\mathit{M}.\mathit{T}$$

Answered by puissant last updated on 18/Oct/21

$$y^{″}=-y\to y^{″}+y=0$$$$y=e^{rx}\to y^{\prime }={re}^{rx}\to y^{″}=r^2{e}^{rx}$$$$y^{″}+y=e^{rx}(r^2+1)=0\to r^2+1=0$$$$\Rightarrow r_1=i=0+i;r_2=-i=0-i$$$$\Rightarrow y=(Acos\left(ix\right)+Bsin\left(ix\right))e^{0x}$$$$\Rightarrow y=(Acos\left(ix\right)+Bsin\left(ix\right))$$

Answered by amin96 last updated on 18/Oct/21

$$\frac{{\mathbf{d}}^2\mathbf{y}}{{\mathbf{dx}}^2}={\mathbf{r}}^2\mathbf{y}=1$$$${\mathbf{r}}^2={\mathbf{i}}^2\Rightarrow \{\begin{array}{l}{\mathbf{r}}_1=\mathbf{i}\\ {\mathbf{r}}_2=-\mathbf{i}\end{array}\Rightarrow \{\begin{array}{l}Im\left\{r\right\}=1=\mathbf{a}\\ Re\left\{r\right\}=0=\mathbf{b}\end{array}$$$$y={\mathbf{c}}_1{x}^0\left(\mathbf{cos}\left(\mathbf{ax}\right)\right)+{\mathbf{c}}_2{x}^0\left(\mathbf{sin}\left(\mathbf{ax}\right)\right)$$$$\mathbf{y}={\mathbf{c}}_1\mathbf{cos}\left(\mathbf{x}\right)+{\mathbf{c}}_2\mathbf{sin}\left(\mathbf{x}\right)$$

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

$$say{y}^{\prime }=\frac{dy}{dx}=u$$$$y^{″}=\frac{du}{dx}=u\frac{du}{dy}$$$$u\frac{du}{dy}=-y$$$$\int udu=-\int ydy$$$$u^2=C^2-y^2$$$$u=\frac{dy}{dx}=\pm \sqrt{C^2-y^2}$$$$\int \frac{dy}{\sqrt{C^2-y^2}}=\pm \int dx$$$${\sin }^{-1}\frac{y}{C}=\pm (x+C_2)$$$$\Rightarrow y=C_1\sin (x+C_2)$$

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