Page 157 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes  (1) becomes
dz 3
 2z.  x , which is linear in z.
dx

I.F.  2  2xdx  e x 2 .

 The solution is
2 2
3
x
ze x   x .e dx c

2
2
x

  x .e .xdx c
1 t 2
  te dt c (t  x )

2
1 t
 e t  1  c
2
2 2
2
ze x  1 x  x   1  c
e
2
1 2  x 2
or z   x   1  ce
2
1 2  x 2
 tan y   x   1  ce
2
which is the required solution.

Example:

dy y 2
Solve   y .
dx x
Solution:
2
Dividing throughout by y , we have
1  dy  1 . 1  1
y 2 dx x y …..(1)

1
Putting  z
y

1 dy dz
  2  .
y dx dx
 Equation (1) becomes

dz 1
 z  1 …..(2)
dx x
which is linear in z.

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