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Question Number 89908 by I want to learn more last updated on 19/Apr/20

$$\mathrm{Solve}\mathrm{the}\mathrm{differential}\mathrm{equstion}:$$$$\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=\frac{{\mathrm{y}}_0-{2\mathrm{y}}_{-1}+{\mathrm{y}}_{-2}}{{\mathrm{h}}^2}$$

Commented by I want to learn more last updated on 19/Apr/20

$$\mathrm{Does}\mathrm{this}\mathrm{question}\mathrm{really}\mathrm{valid}?$$

Commented by MJS last updated on 20/Apr/20

$$\mathrm{what}\mathrm{does}{y}_k\mathrm{mean}?$$$$\mathrm{does}\mathrm{it}\mathrm{mean}{y}_k=f\left(k\right)?$$

Commented by I want to learn more last updated on 20/Apr/20

$$\mathrm{Yes}\mathrm{sir}$$

Commented by I want to learn more last updated on 20/Apr/20

$$\mathrm{Please}\mathrm{help}\mathrm{me}$$

Commented by mr W last updated on 20/Apr/20

$$say\frac{{\mathrm{y}}_0-{2\mathrm{y}}_{-1}+{\mathrm{y}}_{-2}}{{\mathrm{h}}^2}=k=constant$$$$\frac{d^{2}y}{{dx}^2}=k$$$$\Rightarrow y=\frac{{kx}^2}{2}+bx+c$$$$y_0=c$$$$y_{-1}=\frac{k}{2}-b+c$$$$y_{-2}=2k-2b+c$$$$y_0-2y_{-1}+y_{-2}=c-k+2b-2c+2k-2b+c={kh}^2$$$$k={kh}^2$$$$ifh\ne \pm 1:$$$$\Rightarrow k=0$$$$\Rightarrow y=bx+c$$$$ifh=\pm 1:$$$$\Rightarrow k=any=2a$$$$\Rightarrow y={ax}^2+bx+c$$

Commented by I want to learn more last updated on 20/Apr/20

$$\mathrm{Wow},\mathrm{i}\mathrm{appreciate}\mathrm{sir}.$$

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