Question

6. Use Euler's method to approximate the solution to y'= xºy - y at x = 1.2 when y(0) =1. Use a step size of h= .1.

Answer 1

Given y′=x³y-y²,y(0)=1,h=0.1,y(1.2)=?

Euler method
y₁=y₀+hf(x₀,y₀)=1+(0.1)f(0,1)=1+(0.1)⋅(-1)=1+(-0.1)=0.9

y₂=y1+hf(x₁,y₁)=0.9+(0.1)f(0.1,0.9)=0.9+(0.1)⋅(-0.8091)=0.9+(-0.0809)=0.8191

y₃=y₂+hf(x₂,y₂)=0.8191+(0.1)f(0.2,0.8191)=0.8191+(0.1)⋅(-0.6644)=0.8191+(-0.0664)=0.7527

y₄=y3+hf(x₃,y₃)=0.7527+(0.1)f(0.3,0.7527)=0.7527+(0.1)⋅(-0.5462)=0.7527+(-0.0546)=0.698

y₅=y4+hf(x₄,y₄)=0.698+(0.1)f(0.4,0.698)=0.698+(0.1)⋅(-0.4426)=0.698+(-0.0443)=0.6538

y₆=y₅+hf(x₅,y₅)=0.6538+(0.1)f(0.5,0.6538)=0.6538+(0.1)⋅(-0.3457)=0.6538+(-0.0346)=0.6192

y₇=y₆+hf(x₆,y₆)=0.6192+(0.1)f(0.6,0.6192)=0.6192+(0.1)⋅(-0.2497)=0.6192+(-0.025)=0.5942

y₈=y₇+hf(x₇,y₇)=0.5942+(0.1)f(0.7,0.5942)=0.5942+(0.1)⋅(-0.1493)=0.5942+(-0.0149)=0.5793

y₉=y8+hf(x₈,y₈)=0.5793+(0.1)f(0.8,0.5793)=0.5793+(0.1)⋅(-0.039)=0.5793+(-0.0039)=0.5754

y₁₀=y₉+hf(x₉,y₉)=0.5754+(0.1)f(0.9,0.5754)=0.5754+(0.1)⋅(0.0884)=0.5754+(0.0088)=0.5843

y₁₁=y₁₀+hf(x₁₀,y₁₀)=0.5843+(0.1)f(1,0.5843)=0.5843+(0.1)⋅(0.2429)=0.5843+(0.0243)=0.6085

y₁₂=y₁₁+hf(x₁₁,y₁₁)=0.6085+(0.1)f(1.1,0.6085)=0.6085+(0.1)⋅(0.4396)=0.6085+(0.044)=0.6525

∴y(1.2)=0.6525