The value of first iterative root y1 = y(0.1) of y'+ 2y = 0, y(0) = 1, h = 0.1 i
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Answer:
y₁=0.8, y₂=0.64
Step-by-step explanation:
Given y¹+2y=0
we rewrite as y¹=-2y, y(0)=1,
which is of the form of equation f(x,y)=-2y, x₀=0, and y₀=1
Euler's method yileds
y₁=y₀+hf(x₀,y₀)
=1+(0.1)f(0,1)
=1+(0.1)(-2) since f(x,y)=-2y⇒f(0,1)=-2×1=-2×0.8=-1.6
=1+(-0.2)=1-0.2
y₁ =0.8
y₂=y₁+hf(x₁,y₁)
=0.8+(0.1)f(0.1,0.8)
=0.8+0.1×-1.6 since,f(0.1,0.8)=-2
=0.8-0.16
y₂=0.64
So the value y₁=0.8, y₂=0.64
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