Use Fourth-order Runge-Kutta method to find u(0.2), the initial value problem du/dx = –2tu^2 where
u(0)1, using h = 0.1
Answers
Answer:
Answer:
Answer: L=2m,
Answer: L=2m,d=3mm,A=
Answer: L=2m,d=3mm,A= 4
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π
Answer: L=2m,d=3mm,A= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL=
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 4
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm
Answer: L=2m,d=3mm,A= 49π ×10 −6 m 2 ΔL= 49π ×10 −6 ×10 11 30×2 =8.48×10 −5 m=0.085mm .