Page 123 - DMTH202_BASIC_MATHEMATICS_II

Basic Mathematics-II

Notes dy
Example: Solve x 2  y  1
dx
Solution:

dy
y
2
Here x   1
dx
dy dx
or  2
1  y x
Integrating, we get

1
–log (1 – y) = + c
x
1
or log (1 – y) = – c
x
1 1
c
x where e =A.
 1–y = e   x  Ae ; -c
1
or y = x
1  Ae
which is the required solution.

2
Example: Solve (xy + x)dx+(yx +y)dy = 0.
2
Solution:
2
2
Here x(y +1)dx + y (x +1)dy = 0
xdx ydx
Or 2  2  0
x  1 y  1
Integrating, we get

1 1
2
2
log (x + 1) + log (y + 1) = c
2 2
2
or log (x + 1)+ (y + 1) = 2c
2
2
2
2
or (x +1) (y +1) = e e = A, which is the required solution.
Example:
dy 3x 2y 2 2y
Solve  e  x e
dx
Solution:

dy 3x 2y 2 2y
Here  e  x e
dx

2y
or e dy   e 3x  x 2  dx

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