2. a
p(x,t) = 1
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ρ(6,2) is the same value as ρ(4,0)
1.00000000000000 |
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1 |
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2.b
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5.00000000000000 |
-t + x + 1 |
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2.c
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1.00000000000000 |
1 |
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prob 3
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0.000123409804086680 |
e^(-(2*t - x + 2)^2) |
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problem 4
initial condition
ρ(x,t) = 0
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0.000000000000000 |
0 |
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ρ(x,t)=x
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6.00000000000000 |
-2*t + x |
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ρ(x,t)=x^2
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36.0000000000000 |
(2*t - x)^2 |
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ρ(x,t)=e^x
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403.428793492735 |
e^(-2*t + x) |
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