5. Solve the following initial value problem (IVP) '–xy = x y(0) = by using (a) second-order Taylor's series method with h = 0.2 and 0


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5. Solve the following initial value problem (IVP)<br>'–xy = x y(0) =<br>by using<br>(a) second-order Taylor's series method with h = 0.2 and 0<xs1.<br>(b) second-order Taylor's series method with h = 0.1,0.25,0.5 and 0<x<1<br>(c) third-order Taylor's series method with h = 0.2,0.25,0.5 and 0<x<1.<br>Hence, if the exact solution is y = 2eT -1, find its errors.<br>ANSWER:<br>5. (a)<br>h= 0.5<br>error|<br>errot<br>exact<br>exact<br>1.000<br>1.000<br>1.000<br>1.000<br>1<br>0.2<br>1.040<br>1.040<br>0.000<br>0.5<br>1.0<br>1.250<br>2.164<br>1.266<br>2.297<br>0.016<br>0.133<br>1.167<br>1.394<br>0.003<br>0.007<br>0.4<br>1.164<br>1.387<br>0.6<br>4<br>0.8<br>1.0<br>1.738<br>2.266<br>1.754<br>0.016<br>0.031<br>(c) h= 0.2<br>2.297<br>Jerror|<br>exact<br>1.000<br>1.000<br>(b) h =0.1<br>0.2<br>1.040<br>1.166<br>1.393<br>1.040<br>0.000<br>error|<br>exact<br>0.4<br>3<br>0.6<br>1.167<br>0.001<br>0.001<br>0.002<br>0.004<br>1.000<br>1.000<br>1.394<br>1.010<br>1.040<br>1.091<br>1<br>0.1<br>1.010<br>0.000<br>4<br>0.8<br>1.752<br>1.754<br>0.2<br>3<br>0.3<br>1.040<br>1.092<br>1.167<br>0.000<br>0.001<br>1.0<br>2.293<br>2.297<br>4<br>0.4<br>1.165<br>0.002<br>h = 0.25<br>0.5<br>0.6<br>1.264<br>1.391<br>1.266<br>1.394<br>0.002<br>0.003<br>Jerrot|<br>exact<br>1.000<br>1.000<br>7<br>0.7<br>0.8<br>1.551<br>1.749<br>1.991<br>1.555<br>1.754<br>1.999<br>0.004<br>0.005<br>0.25<br>0.50<br>0.75<br>1.062<br>1.263<br>1.643<br>1.063<br>1.266<br>1.650<br>0.001<br>0.003<br>0.007<br>0.9<br>0.008<br>10 1.0<br>2.287<br>2.297<br>0.010<br>3<br>4<br>1.00<br>2.285<br>2.297<br>0.012<br>h-0.25<br>h= 0.5<br>errot|<br>exact<br>Jerrot|<br>X,<br>exact<br>1.000<br>1.000<br>1<br>0.25<br>1.062<br>1.063<br>0.001<br>1.000<br>1.000<br>1<br>0.5<br>1.0<br>1.250<br>2.241<br>1.266<br>2.297<br>0.50<br>1.259<br>1.266<br>0.007<br>0.016<br>3<br>0.75<br>1.629<br>1.650<br>0.021<br>0.056<br>4<br>1.00<br>2.249<br>2.297<br>0.048<br>12<br>123<br>in<br>no790 a<br>

Extracted text: 5. Solve the following initial value problem (IVP) '–xy = x y(0) = by using (a) second-order Taylor's series method with h = 0.2 and 0<><1 (c)="" third-order="" taylor's="" series="" method="" with="" h="0.2,0.25,0.5" and=""><><1. hence,="" if="" the="" exact="" solution="" is="" y="2eT" -1,="" find="" its="" errors.="" answer:="" 5.="" (a)="" h="0.5" error|="" errot="" exact="" exact="" 1.000="" 1.000="" 1.000="" 1.000="" 1="" 0.2="" 1.040="" 1.040="" 0.000="" 0.5="" 1.0="" 1.250="" 2.164="" 1.266="" 2.297="" 0.016="" 0.133="" 1.167="" 1.394="" 0.003="" 0.007="" 0.4="" 1.164="" 1.387="" 0.6="" 4="" 0.8="" 1.0="" 1.738="" 2.266="" 1.754="" 0.016="" 0.031="" (c)="" h="0.2" 2.297="" jerror|="" exact="" 1.000="" 1.000="" (b)="" h="0.1" 0.2="" 1.040="" 1.166="" 1.393="" 1.040="" 0.000="" error|="" exact="" 0.4="" 3="" 0.6="" 1.167="" 0.001="" 0.001="" 0.002="" 0.004="" 1.000="" 1.000="" 1.394="" 1.010="" 1.040="" 1.091="" 1="" 0.1="" 1.010="" 0.000="" 4="" 0.8="" 1.752="" 1.754="" 0.2="" 3="" 0.3="" 1.040="" 1.092="" 1.167="" 0.000="" 0.001="" 1.0="" 2.293="" 2.297="" 4="" 0.4="" 1.165="" 0.002="" h="0.25" 0.5="" 0.6="" 1.264="" 1.391="" 1.266="" 1.394="" 0.002="" 0.003="" jerrot|="" exact="" 1.000="" 1.000="" 7="" 0.7="" 0.8="" 1.551="" 1.749="" 1.991="" 1.555="" 1.754="" 1.999="" 0.004="" 0.005="" 0.25="" 0.50="" 0.75="" 1.062="" 1.263="" 1.643="" 1.063="" 1.266="" 1.650="" 0.001="" 0.003="" 0.007="" 0.9="" 0.008="" 10="" 1.0="" 2.287="" 2.297="" 0.010="" 3="" 4="" 1.00="" 2.285="" 2.297="" 0.012="" h-0.25="" h="0.5" errot|="" exact="" jerrot|="" x,="" exact="" 1.000="" 1.000="" 1="" 0.25="" 1.062="" 1.063="" 0.001="" 1.000="" 1.000="" 1="" 0.5="" 1.0="" 1.250="" 2.241="" 1.266="" 2.297="" 0.50="" 1.259="" 1.266="" 0.007="" 0.016="" 3="" 0.75="" 1.629="" 1.650="" 0.021="" 0.056="" 4="" 1.00="" 2.249="" 2.297="" 0.048="" 12="" 123="" in="" no790="">
Formula<br>Second Order Taylor Series Method<br>y(x) = y(x;) + hy'(x;) +

Extracted text: Formula Second Order Taylor Series Method y(x) = y(x;) + hy'(x;) + "(x,) When the Taylor's series is truncated after three terms, it is called second order Taylor's series method. Else, we write as Vis = y; + hy + 2!

Jun 04, 2022
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