e4y (1)
Separating the variables, we get
edy4 y
= e3x dx
Therefore
e4 y
dy
∫−
=
3x
e dx
∫
Fig 9 6
© NCERT
not to be republished
DIFFERENTIAL EQUATIONS
417
or
4
4
y
e−
−
 =
3
C
3
ex
+
or
4 e3x + 3 e– 4y + 12 C = 0 (2)
Substituting x = 0 and y = 0 in (2), we get
4 + 3 + 12 C = 0 or C = 
127
−
Substituting the value of C in equation (2), we get
4e3x + 3e– 4y – 7 = 0,
which is a particular solution of the given differential equation