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Solved Problems

The advantage gained by this step is that solutions for each of these equations can
be easily obtained:

1 9
5
v(x, t) = − cos 2t, w(x, t) = − x .
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Thus,
1 9
5
u(x, t) = − cos 2t − x .
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There are many other solutions. For example, it is easy to check that if we add to
1

the solution above a function of the form f 2x − t , where f(s) is an arbitrary
3
twice differentiable function, a new solution is obtained.
Exercise 4. Solve the equation

2
2
(x + y )u + 2xyu = 0.
x
y
The system of differential equations has the following form
dx dy du
= = .
2
x + y 2 2xy 0

Using the proportionality property, we rewrite the first equation in the form
dx dy dx + dy dx − dy
= , = ,
2
2
2
2
2
x + y 2 2xy x + y + 2xy x + y − 2xy
or
d(x + y) d(x − y)
= .
(x + y) 2 (x − y) 2
Integrating the last equation, we obtain
1 1 1 1 2y y
− = − + C; − = C; = C, = C .
1
2
2
x + y x − y x − y x + y x − y 2 x − y 2
The second equation of the system du = 0 has the solution u = C . So, a general
2
solution of the initial equation is following

y y
Φ , u = 0, u = φ .
2
2
x − y 2 x − y 2
Exercise 5. Find the solution of the equation
u
2
u − = x y
x
x
satisfying the condition
2
u(1, y) = y .

Then
u du dx
u − = 0, = , ln u = ln x + ln C(y), u(x, y) = xC(y).
x
x u x

5