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Question Number 89908 by I want to learn more last updated on 19/Apr/20
Solve the differential equstion:        (d^2 y/dx^2 )   =   ((y_0   −  2y_(−1)   +  y_(−2) )/h^2 )
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equstion}: \\ $$$$\:\:\:\:\:\:\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }\:\:\:=\:\:\:\frac{\mathrm{y}_{\mathrm{0}} \:\:−\:\:\mathrm{2y}_{−\mathrm{1}} \:\:+\:\:\mathrm{y}_{−\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} } \\ $$
Commented by I want to learn more last updated on 19/Apr/20
Does this question really valid?
$$\mathrm{Does}\:\mathrm{this}\:\mathrm{question}\:\mathrm{really}\:\mathrm{valid}? \\ $$
Commented by MJS last updated on 20/Apr/20
what does y_k  mean?  does it mean y_k =f(k)?
$$\mathrm{what}\:\mathrm{does}\:{y}_{{k}} \:\mathrm{mean}? \\ $$$$\mathrm{does}\:\mathrm{it}\:\mathrm{mean}\:{y}_{{k}} ={f}\left({k}\right)? \\ $$
Commented by I want to learn more last updated on 20/Apr/20
Yes sir
$$\mathrm{Yes}\:\mathrm{sir} \\ $$
Commented by I want to learn more last updated on 20/Apr/20
Please help me
$$\mathrm{Please}\:\mathrm{help}\:\mathrm{me} \\ $$
Commented by mr W last updated on 20/Apr/20
say ((y_0   −  2y_(−1)   +  y_(−2) )/h^2 )=k=constant  (d^2 y/dx^2 )=k  ⇒y=((kx^2 )/2)+bx+c  y_0 =c  y_(−1) =(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   if h≠±1:  ⇒k=0  ⇒y=bx+c  if h=±1:  ⇒k=any=2a  ⇒y=ax^2 +bx+c
$${say}\:\frac{\mathrm{y}_{\mathrm{0}} \:\:−\:\:\mathrm{2y}_{−\mathrm{1}} \:\:+\:\:\mathrm{y}_{−\mathrm{2}} }{\mathrm{h}^{\mathrm{2}} }={k}={constant} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }={k} \\ $$$$\Rightarrow{y}=\frac{{kx}^{\mathrm{2}} }{\mathrm{2}}+{bx}+{c} \\ $$$${y}_{\mathrm{0}} ={c} \\ $$$${y}_{−\mathrm{1}} =\frac{{k}}{\mathrm{2}}−{b}+{c} \\ $$$${y}_{−\mathrm{2}} =\mathrm{2}{k}−\mathrm{2}{b}+{c} \\ $$$${y}_{\mathrm{0}} −\mathrm{2}{y}_{−\mathrm{1}} +{y}_{−\mathrm{2}} ={c}−{k}+\mathrm{2}{b}−\mathrm{2}{c}+\mathrm{2}{k}−\mathrm{2}{b}+{c}={kh}^{\mathrm{2}} \\ $$$${k}={kh}^{\mathrm{2}} \\ $$$${if}\:{h}\neq\pm\mathrm{1}: \\ $$$$\Rightarrow{k}=\mathrm{0} \\ $$$$\Rightarrow{y}={bx}+{c} \\ $$$${if}\:{h}=\pm\mathrm{1}: \\ $$$$\Rightarrow{k}={any}=\mathrm{2}{a} \\ $$$$\Rightarrow{y}={ax}^{\mathrm{2}} +{bx}+{c} \\ $$
Commented by I want to learn more last updated on 20/Apr/20
Wow,  i appreciate sir.
$$\mathrm{Wow},\:\:\mathrm{i}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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