, the solution obtained from the general
solution by giving particular values to the arbitrary constants is called a particular
solution of the differential equation Example 2 Verify that the function y = e– 3x is a solution of the differential equation
2
2
6
0
d y
dy
y
dx
dx
+
−
=
Solution Given function is  y = e– 3x Differentiating both sides of equation with respect
to x , we get
3
3
x
dy
e
dx
−
= − (1)
Now, differentiating (1) with respect to x, we have
2
d y2
dx
 = 9 e – 3x
Substituting the values of  
2
2 ,
d y dy
dx
dx
and y in the given differential equation, we get
L