K01 koz = k03 K04 = = = P(1) ≈ P₁ = Iteration 2: K11 = K12 K13 = K14 = = P(2) ≈ P₂ =
K01 koz = k03 K04 = = = P(1) ≈ P₁ = Iteration 2: K11 = K12 K13 = K14 = = P(2) ≈ P₂ =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
Expert Solution
Step 1
Given P′=0.05P, P(0)=100, h=1
4th order Runge-Kutta method
k01=hP(t0,y0)=(1)P(0,100)=(1)⋅(5)=5
k02=hP(t0+h/2,P0+k01/2)=(1)P(0.5,102.5)=(1)⋅(5.125)=5.125
k03=hP(t0+h/2,P0+k02/2)=(1)P(0.5,102.5625)=(1)⋅(5.1281)=5.1281
k04=hP(t0+h,P0+k03)=(1)P(1,105.1281)=(1)⋅(5.2564)=5.2564
P1=P0+16(k01+2k02+2k03+k04)
P1=100+16[5+2(5.125)+2(5.1281)+(5.2564)]
P1=105.1271
∴P(1)=105.1271
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