Kurs_vysshei_matematiki_UP_Berkov_N.A._2007-2
.pdf
a) |
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dt |
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du |
du = 0 |
u = C1; |
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1 |
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b) |
dt |
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dx |
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dx |
= 2et − x; |
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1 |
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2et − x |
dt |
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dx |
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+ x = 2et. |
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dt |
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dxdt + x = 0 dxdt = −x dxx = −dt
ln x − ln C = −t ln Cx = −t x = Ce−t.
C 
C(t)
x 
C(t) 
x = C(t)e−t |
dx |
+ C(t)e−t = |
dC(t) |
e−t. |
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dt |
dt |
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dC(t) |
e−t − C(t)e−t = 2et − C(t)e−t |
dC(t) |
= 2et |
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dt |
dt |
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dC(t) = 2e2tdt C(t) = e2t + C2 x = (e2t + C2)e−t |
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x = et + C2e−t C2 = xet − e2t. |
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u x |
t |
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C1 = u C2 = |
= xet − e2t t = 0 u = x |
C1 = C2 + 1 |
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u(x, t) = xet − e2t + 1.
a 
u
x 
h 
tau
h 
tau
t
x
x
B
C
h
D
= 
L
D
∂u
B
= 
U 
t
U 
Ctrl 
Shif t 
B
t 
Y 
Ox 
Oy 
Ox
X 
OY
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u |
t = 1 |
B17 |
= B3 EXP ($A$14) − EXP (2 |
$A$14) + 1 |
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L17 |
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t = 1 |
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h |
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0, 1 |
τ |
0, 1 |
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t\x |
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t = 1
x 
x 
h 
τ 
B
D
U 
y
n
ynp1 
n + 1
tauP r 
Ox 
ynp1 
M P r 
u(t, x) 
t = 1