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Example 7.2 To find solutions of the system of equations .
 dy
 dx  2y z  sin ,x
 dz
  4y  2z  cos x .
 dx

 We differentiate the first equation on x :
2
d y dy dz
.  2   cos x (7.6)
dx 2 dx dx

From the first equation of the given system we determine
dy
z : z  sin x  2y  . We put this value in the second
dx
equation of the system. Obsessed:

dz dy
 4y  2sin x  4y  2  cos x
dx dx
or
dz dy
 cos x  2sin x  2 .
dx dx
dz
We put the found value in equation (7.6) :
dx
2
d y dy dy
 2  cos x  2sin x  2  cos x
dx 2 dx dx
or
2
d y
  2sin x .
dx 2
It is equation of the second order. Integrating him twice, we
find
y  2sin x C x C  .
1 2
Putting y and y into equation (6.2), we will find
z  sin x  4sin x  2C x  2C  2cos x C 
1 2 1
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