Question Number 63152 by Rio Michael last updated on 29/Jun/19
$${The}\:{Variables}\:{x}\:{and}\:{y}\:{satisfy}\:{the}\:{differential}\:{equation}\: \\ $$$$\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−{x}\frac{{dy}}{{dx}}\:+\:{y}\:=\:{x}^{\mathrm{2}} \:\:{use}\:{the}\:{approximations} \\ $$$$\:\:\:\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)_{{n}} \approx\:\frac{{y}_{{n}+\mathrm{1}} −\mathrm{2}{y}_{{n}} +{y}_{{n}−\mathrm{1}} }{{h}^{\mathrm{2}\:} }\:{and}\:\:\left(\frac{{dy}}{{dx}}\right)\approx\frac{{y}_{{n}+\mathrm{1}} −{y}_{{n}−\mathrm{1}} }{\mathrm{2}{h}\:}\:{to}\:{show}\:{that} \\ $$$$\:\:\left(\mathrm{2}−{hx}_{{n}} \right){y}_{{n}+\mathrm{1}} \approx\mathrm{2}{x}_{{n}} ^{\mathrm{2}} \:+\:\mathrm{398}{y}_{{n}} −\left(\mathrm{200}\:+\:\mathrm{10}{x}_{{n}} \right){y}_{{n}−\mathrm{1}} \\ $$$${given}\:{that}\:{y}=\mathrm{1}\:{and}\:{x}=\mathrm{0}\:{and}\:{that}\:{y}_{−\mathrm{1}} ={y}_{\mathrm{1}} \:{show}\:{that}\:{y}_{\mathrm{1}} =\mathrm{0}.\mathrm{995} \\ $$$${Evaluate}\:{y}\:{when}\:{x}=\mathrm{0}.\mathrm{3},\:{giving}\:{your}\:{answercorrect}\:{to}\:\mathrm{3}\:{decimal}\:{places} \\ $$