Solution Given that y = 3e2x + 2e3x Then
dy
dx  = 6e2x + 6e3x = 6 (e2x + e3x)
Therefore
2
2
d y
dx
 = 12e2x + 18e3x = 6 (2e2x + 3e3x)
Hence
2
2
5
d y
dy
dx
dx
−
 + 6y = 6 (2e2x + 3e3x)
– 30 (e2x + e3x) + 6 (3e2x + 2e3x) = 0
Rationalised 2023-24
CONTINUITY AND DIFFERENTIABILITY
139
Example 38 If y = sin–1 x, show that (1 – x2) 
2
2
0
d y
dy
x dx
dx
−
= Solution We have y = sin–1x Then
dy
dx  = 
2
1
(1
x)
−
or
2
(1
)
1
dy
x
dx
−
=
So
2
(1
)