Methods of Finding Particular Integral
There are different methods of finding the particular integral. The method to be applied depends on the type of function involved. There is a general method which can be applied to any kind of question but it is comparatively lengthy. The short methods are also there which are specific according to the type of function involved.
General Method
If both m1 and m2 are constants, the expressions (D – m1) (D – m2) y and (D– m2) (D – m1) y are equivalent i.e. the expression is independent of the order of operational factors.
We will explain the method with the help of following
Example: Solve d2y/ dx2 – 5 dy/dx +6y = e3x.
Solution:
The equation can be written as
(D2 – 5D + 6)y = e3x
(D – 3) (D – 2)y = e3x
C.F. = c1 e3x + c2e2x
And P.I. =1/ (D-3). 1/ (D-2) e3x
= 1/ (D-3) e2x ∫ e3x e-2xdx
=1/ (D-3) e2x ex
= e3x ∫ e3x e-3x dx = x.e3x
y = c1e3x +c2e2x +xe3x
P.I. can be found by resolving
1/f (D) = 1/ (D-3). 1/ (D-2)
Now using partial fractions,
1/f (D) = 1/ (D-3). 1/ (D-2)
= 1/ (D-3) – 1/ (D-2)
Hence, the required P.I. is [1(D-3) – 1/ (D-2)] e3x
= 1/ (D-3) e3x – 1/ (D-2) e3x
= e3x ∫ e3x e-3x dx – e2x ∫ e3x e-2x dx
= xe3x – e3x
Second term can be neglected as it is included in the first term of the C.F.
Short Method of Finding P.I.
In certain cases, the P.I. can be obtained by methods shorter than the general method.
(i). To find P.I. when X = eax in f(D) y = X, where a is constant
y = 1/f(D)
1/f(D) eax = 1/f(a) eax , if f(a) ≠ 0.
1/f(D) eax = xr/f r(a) eax , if f(a) = 0, where f(D) = (D-a)rf(D)
Example: Solve (D3 – 5D2 + 7D – 3)y = e3x.
Solution:
(D – 1)2 (D – 3) y = e3x
C.F. = aex + bx ex + ce3x
And P.I. = 1/ (D3-5D2+7D-3) e3x
= x/ (3D2-10D+7) e3x
= ¼ xe3x
y = aex + bx ex + ce3x + ¼ xe3x.
(ii). To find P.I. when X = cos ax or sin ax
f (D) y = X
y = 1/f(D) sinax
If f (– a2) ≠ 0 then 1/f(D2) sinax = 1/f(-a2) sinax
If f (– a2) = 0 then (D2 + a2) is at least one factor of f (D2)
Let f (D2) = (D2 + a2)r f (D2)
Where f (– a2) ≠ 0
[1/f (D2)] sin ax = 1/ (D2+a2). 1/φ(D2) sin ax
= 1/ φ(-a2). 1/ (D2+a2) sin ax
Now, when r=1, 1/(D2 + a2) sin ax = -x/2a. cos ax
Similarly If f(–a)2 ≠ 0 then 1/f(D2) cos ax = 1/f(-a2) cosax
And 1/ (D2+a2) cos ax = x/2a .sin ax
Example: Solve (D2 – 5D + 6) y = sin3x.
Solution: We can factorize the given term as (D – 2) (D – 3)y = sin3x
C.F. = ae2x + be3x
P.I. = 1/ (D2 -5D +6) sin 3x
= 1/ [(-9) -5D +6] sin 3x
= 1/ (-5D-3) sin 3x
= - (5D-3). 1/ (25D2 -9) sin3x
= 1/234 (15 cos 3x-3 sin3x)
= ae2x + be3x + 1/234 (15 cos 3x-3 sin3x)
(iii) To find the P.I. when X = xm where m ∈ N
f (D) y = xm
y = 1/ f(D) xm
We will explain the method by taking an example
Example: Find P.I. of (D3 + 3D2 + 2D) y = x2.
Solution: The P.I. is given by 1/ (D3+3D2+2D) x2
= 
= 
= 
= 1/2D (1-3D/2+7/4D2 + ….) x2
= 1/2D (x2-3x +7/2)
And 1/2D (x2-3x+7/2) = 1/2(x3/3-3x2/2+7/2x) = 1/12 (2x3 -9x2+21x).
(iv) To find the value of 1/f(D) eax V where ‘a’ is a constant and V is a function of x
1/f (D) .eax V = eax.1/f (D+a). V
Example: Solve (D2 + 2) y = x2 e3x.
Solution: C.F. = a cos √2x + b sin √2x
P.I. = 1/ (D2+2). x2e3x = e3x. 1/ [(D+3)2 +2] .x2
= e3x. 1/(D2 +6D+11). x2
= 1/11. e3x (1+ (6D+D2)/11)-1 x2
= 1/11 e3x [1- (6D+D2)/11 + 36/121 . D2 +…) x2
= 1/11 e3x (1-6D/11+25/121 . D2+…) x2
= 1/11 e3x (x2 – 12/11 x + 50/121)
= a cos √2x + b sin √2x + e3x/11 (x2-12/11 x +50/121)
(v). To find 1/f (D). xV where V is a function of x
1/f (D). xV = [x- 1/f(D). f'(D)] 1/f(D) V
Watch this Video for more reference
Example: Solve (D2 + 4) y = x sin2x.
Solution: C.F. = c1 cos 2x + c2 sin2x
P.I. = 1/ (D2+4). x sin 2x
= {x- 1/ (D2+4) .2D}. 1/ (D2+4). sin 2x
= {x- 1/ (D2+4). 2D} {-x/4 cos 2x}
= -x2/4 cos 2x + ½.1/(D2+4) (cos 2x- 2x sin 2x)
= -x2/4 .cos 2x + ½ 1/(D2+4) cos 2x – 1/(D2+4) x sin 2x
= -x2/8 cos 2x +1/16 x sin 2x
So, y= c1 cos 2x + c2 sin 2x – x2/ 8 cos 2x + 1/16 x sin 2x.
Some Results on Tangents and Normals:
(i) The equation of the tangent at P(x, y) to the curve y= f(x)
is Y – y = dy/dx(X-x)
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(ii) The equation of the normal at point P(x, y) to the curve y = f(x) is Y – y = [-1/ (dy/dx) ].(X – x ) (iii) The length of the tangent = CP = y √[1+(dx/dy)2] (iv) The length of the normal = PD = y √[1+(dy/dx)2] |
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(v) The length of the Cartesian subtangent = CA = y dy/dx
(vi) The length of the Cartesian subnormal = AD = y dy/dx
(viii) The initial ordinate of the tangent = OB = y – x.dy/dx
The methods of finding P.I. are quite simple and with a bit of practice they can easily fetch you 2-3 questions in the JEE. You may also consult the Sample Papers to get an idea about the types of questions asked.
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