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Arnold's Trivium
1991 Arnold's Trivium
78
Problem 78
Problem 78
Source:
July 1, 2010
trigonometry
Problem Statement
Solve the Cauchy problem
∂
2
A
∂
t
2
=
9
∂
2
A
∂
x
2
−
2
B
,
∂
2
B
∂
t
2
=
6
∂
2
B
∂
x
2
−
2
A
\frac{\partial ^2A}{\partial t^2}=9\frac{\partial^2 A}{\partial x^2}-2B,\;\frac{\partial^2 B}{\partial t^2}=6\frac{\partial^2 B}{\partial x^2}-2A
∂
t
2
∂
2
A
=
9
∂
x
2
∂
2
A
−
2
B
,
∂
t
2
∂
2
B
=
6
∂
x
2
∂
2
B
−
2
A
A
∣
t
=
0
=
cos
x
,
B
∣
t
=
0
=
0
,
∂
A
∂
t
∣
t
=
0
=
∂
B
∂
t
∣
t
=
0
=
0
A|_{t=0}=\cos x,\; B|_{t=0}=0,\; \left.\frac{\partial A}{\partial t}\right|_{t=0}=\left.\frac{\partial B}{\partial t}\right|_{t=0}=0
A
∣
t
=
0
=
cos
x
,
B
∣
t
=
0
=
0
,
∂
t
∂
A
t
=
0
=
∂
t
∂
B
t
=
0
=
0
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