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Complementary Function in BSc Maths

November 12, 2026 Differential Equation(English) No Comments
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1 1.Complementary Function in BSc Maths
1.1 2.Complementary Function in BSc Maths Examples:
1.2 3.Practice Problems of Complementary Function in BSc Maths for Students
1.2.1 4.Frequently Asked Questions Related to Complementary Function in BSc Maths:
1.2.2 5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):
1.2.3 Q: 1. What is complementary function?
1.2.4 Q: 2. What is called a particular integral?
1.2.5 Q: 3. What is meant by auxiliary equation?

1.Complementary Function in BSc Maths

In this article we will learn about “Complementary Function in BSc Maths” we will find out complementary function of differential equation.

Also Read This Article:-  Ordinary differential equation

2.Complementary Function in BSc Maths Examples:

Solve following differential equations:
Example: 1. \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0
Solution: \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 5y = 0
Given equation can be written as:
(D^2 + 2D + 5)y = 0
Auxiliary Equation will be:
m^2 + 2m + 5 = 0
\implies m = \frac{-2 \pm \sqrt{(2)^2 - 4 \times 1 \times 5}}{2 \times 1}
= \frac{-2 \pm \sqrt{4 - 20}}{2}
= \frac{-2 \pm \sqrt{-16}}{2}
= \frac{-2 \pm 4i}{2}
= \frac{2(-1 \pm 2i)}{2}
\implies m = -1 \pm 2i
\text{C.F.} = e^{-x}(C_1 \cos(2x) + C_2 \sin(2x))
Example: 2. \frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} + 4y = 0
Solution: \frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} + 4y = 0
Given equation can be written as:
(D^3 - 3D^2 + 4)y = 0
Auxiliary Equation will be:
m^3 - 3m^2 + 4 = 0
\implies m^3 + m^2 - 4m^2 - 4m + 4m + 4 = 0
\implies m^2(m + 1) - 4m(m + 1) + 4(m + 1) = 0
\implies (m + 1)(m^2 - 4m + 4) = 0
\implies (m + 1)(m^2 - 2m - 2m + 4) = 0
\implies (m + 1)[m(m - 2) - 2(m - 2)] = 0
\implies (m + 1)(m - 2)^2 = 0
\implies m = -1, 2, 2
\text{C.F.} = C_1 e^{-x} + (C_2 + C_3 x)e^{2x}
General solution:
y = C_1 e^{-x} + (C_2 + C_3 x)e^{2x}
Example: 3. \frac{d^3y}{dx^3} - 13\frac{dy}{dx} + 12y = 0
Solution: \frac{d^3y}{dx^3} - 13\frac{dy}{dx} + 12y = 0
Given equation can be written as:
(D^3 - 13D + 12)y = 0
Auxiliary Equation will be:
m^3 - 13m + 12 = 0
\implies m^3 - m^2 + m^2 - m - 12m + 12 = 0
\implies m^2(m - 1) + m(m - 1) - 12(m - 1) = 0
\implies (m - 1)\left[m^2 + 4m - 3m - 12\right] = 0
\implies (m - 1)\left[m(m + 4) - 3(m + 4)\right] = 0
\implies (m - 1)(m - 3)(m + 4) = 0
\implies m = 1, 3, -4
\text{C.F.} = C_1 e^x + C_2 e^{3x} + C_3 e^{-4x}
General solution:
y = C_1 e^x + C_2 e^{3x} + C_3 e^{-4x}
Example: 4. \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0
Solution: \frac{d^3y}{dx^3} - 4\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0
Given equation can be written as:
(D^3 - 4D^2 + 5D - 2)y = 0
Auxiliary Equation will be:
m^3 - 4m^2 + 5m - 2 = 0
\implies m^3 - m^2 - 3m^2 + 3m + 2m - 2 = 0
\implies m^2(m - 1) - 3m(m - 1) + 2(m - 1) = 0
\implies (m - 1)(m^2 - 3m + 2) = 0
\implies (m - 1)(m^2 - m - 2m + 2) = 0
\implies (m - 1)\left[m(m - 2) - 1(m - 2)\right] = 0
\implies (m - 1)(m - 1)(m - 2) = 0
\implies m = 1, 1, 2
\text{C.F.} = (C_1 + C_2 x)e^x + C_3 e^{2x}
General Solution:
y = (C_1 + C_2 x)e^x + C_3 e^{2x}
Example: 5. \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0
Solution: \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0
Given equation can be written as:
(D^2 - 2D + 1)y = 0
Auxiliary Equation will be:
m^2 - 2m + 1 = 0
\implies (m - 1)^2 = 0
\implies m = 1, 1
\text{C.F.} = (C_1 + C_2 x)e^x
General solution:
y = (C_1 + C_2 x)e^x
Example: 6. (D^2 + 5D + 4)y = 0
Solution: (D^2 + 5D + 4)y = 0
Auxiliary Equation will be:
m^2 + 5m + 4 = 0
\implies m^2 + 4m + m + 4 = 0
\implies m(m + 4) + 1(m + 4) = 0
\implies (m + 1)(m + 4) = 0
\implies m = -1, -4
\text{C.F.} = C_1 e^{-x} + C_2 e^{-4x}
General solution:
y = C_1 e^{-x} + C_2 e^{-4x}
Example: 7. (D^3 + 3D^2 + 3D + 1)y = 0
Solution: (D^3 + 3D^2 + 3D + 1)y = 0
Auxiliary Equation will be:
m^3 + 3m^2 + 3m + 1 = 0
\implies m^3 + m^2 + 2m^2 + 2m + m + 1 = 0
\implies m^2(m + 1) + 2m(m + 1) + 1(m + 1) = 0
\implies (m + 1)(m^2 + 2m + 1) = 0
\implies (m + 1)(m + 1)^2 = 0
\implies (m + 1)^3 = 0
\implies m = -1, -1, -1
\text{C.F.} = (C_1 + C_2 x + C_3 x^2)e^{-x}

Example: 8. [(D + 2)(D - 1)^3]y = 0
Solution: [(D + 2)(D - 1)^3]y = 0
Auxiliary Equation will be:
(m + 2)(m - 1)^3 = 0
\implies m = -2, 1, 1, 1
\text{C.F.} = C_1 e^{-2x} + (C_2 + C_3 x + C_4 x^2)e^x
General solution:
y = C_1 e^{-2x} + (C_2 + C_3 x + C_4 x^2)e^x
Example: 9. \frac{d^4y}{dx^4} - 2\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0
Solution: \frac{d^4y}{dx^4} - 2\frac{d^3y}{dx^3} + 2\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 0
Given equation can be written as:
(D^4 - 2D^3 + 2D^2 - 2D + 1)y = 0
Auxiliary Equation will be:
m^4 - 2m^3 + 2m^2 - 2m + 1 = 0
\implies m^4 - m^3 - m^3 + m^2 + m^2 - m - m + 1 = 0
\implies m^3(m - 1) - m^2(m - 1) + m(m - 1) - 1(m - 1) = 0
\implies (m - 1)(m^3 - m^2 + m - 1) = 0
\implies (m - 1)\left[m^2(m - 1) + 1(m - 1)\right] = 0
\implies (m - 1)(m - 1)(m^2 + 1) = 0
\implies m = 1, 1, \pm i
\text{C.F.} = (C_1 + C_2 x)e^x + C_3 \cos x + C_4 \sin x
General solution:
y = (C_1 + C_2 x)e^x + C_3 \cos x + C_4 \sin x
Example: 10. (D^4 + 8D^2 + 16)y = 0
Solution: (D^4 + 8D^2 + 16)y = 0
Auxiliary Equation will be:
m^4 + 8m^2 + 16 = 0
\implies (m^2 + 4)^2 = 0
\implies m = \pm 2i, \pm 2i
\text{C.F.} = (C_1 + C_2 x)\cos 2x + (C_3 + C_4 x)\sin 2x
General solution:
y = (C_1 + C_2 x)\cos 2x + (C_3 + C_4 x)\sin 2x
Example: 11. [(D^2 + D + 1)^2(D - 2)]y = 0
Solution: [(D^2 + D + 1)^2(D - 2)]y = 0
Auxiliary equation will be:
(m^2 + m + 1)^2(m - 2) = 0
m = \frac{-1 \pm \sqrt{(1)^2 - 4 \times 1 \times 1}}{2 \times 1}, \quad m = 2
m = \frac{-1 \pm \sqrt{3}i}{2}, \frac{-1 \pm \sqrt{3}i}{2}, 2
\text{C.F.} = e^{-\frac{x}{2}}\left[(C_1 + C_2 x)\cos\left(\frac{\sqrt{3}}{2}x\right) + (C_3 + C_4 x)\sin\left(\frac{\sqrt{3}}{2}x\right)\right] + C_5 e^{2x}
General solution:
y = e^{-\frac{x}{2}}\left[(C_1 + C_2 x)\cos\left(\frac{\sqrt{3}}{2}x\right) + (C_3 + C_4 x)\sin\left(\frac{\sqrt{3}}{2}x\right)\right] + C_5 e^{2x}
Example: 12. (D^2 + 1)y = 0, given that y = 2, when x = 0 and y = -2 when x = \frac{\pi}{2}.
Solution: (D^2 + 1)y = 0
Auxiliary Equation will be:
m^2 + 1 = 0 \implies m = \pm i
\text{C.F.} = C_1 \cos x + C_2 \sin x
General solution:
y = C_1 \cos x + C_2 \sin x
Given y = 2, when x = 0:
2 = C_1 \cos 0 + C_2 \sin 0 \implies C_1 = 2
y = -2 when x = \frac{\pi}{2}:
-2 = C_1 \cos\frac{\pi}{2} + C_2 \sin\frac{\pi}{2}
-2 = C_2
y = -2\sin x + 2\cos x
= 2(\cos x - \sin x)
= 2\sqrt{2}\left(\frac{1}{\sqrt{2}}\cos x - \frac{1}{\sqrt{2}}\sin x\right)
= 2\sqrt{2}\left(\cos\frac{\pi}{4}\cos x - \sin\frac{\pi}{4}\sin x\right)
\implies y = 2\sqrt{2}\cos\left(x + \frac{\pi}{4}\right)
Example: 13. \frac{d^2x}{dt^2} + \mu x = 0, \mu > 0 given that x = a and \frac{dx}{dt} = 0 when t = \frac{\pi}{\sqrt{\mu}}.
Solution: \frac{d^2x}{dt^2} + \mu x = 0
Given equation can be written as:
(D^2 + \mu)x = 0
Auxiliary equation will be:
m^2 + \mu = 0 \implies m = \pm \sqrt{\mu}i
\text{C.F.} = C_1 \cos(\sqrt{\mu}t) + C_2 \sin(\sqrt{\mu}t)
General solution:
x = C_1 \cos(\sqrt{\mu}t) + C_2 \sin(\sqrt{\mu}t)
\frac{dx}{dt} = -C_1 \sqrt{\mu} \sin(\sqrt{\mu}t) + \sqrt{\mu} C_2 \cos(\sqrt{\mu}t)
Given x = a when t = \frac{\pi}{\sqrt{\mu}}:
a = C_1 \cos\left(\sqrt{\mu} \times \frac{\pi}{\sqrt{\mu}}\right) + C_2 \sin\left(\sqrt{\mu} \times \frac{\pi}{\sqrt{\mu}}\right)
a = C_1 \cos\pi + C_2 \sin\pi
a = C_1(-1) + C_2 \times 0 \implies C_1 = -a
\frac{dx}{dt} = 0 when t = \frac{\pi}{\sqrt{\mu}}:
0 = -C_1 \sqrt{\mu} \sin\left(\sqrt{\mu} \times \frac{\pi}{\sqrt{\mu}}\right) + \sqrt{\mu} C_2 \cos\left(\sqrt{\mu} \times \frac{\pi}{\sqrt{\mu}}\right)
0 = -C_1 \sqrt{\mu} \sin\pi + \sqrt{\mu} C_2 \cos\pi
0 = \sqrt{\mu} C_2 (-1) \implies C_2 = 0
x = -a \cos(\sqrt{\mu}t)
Example: 14. \frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0
When R^2C = 4L and R, C, L are constant.
Solution: \frac{d^2i}{dt^2} + \frac{R}{L}\frac{di}{dt} + \frac{i}{LC} = 0
Given equation can be written as:
\left(D^2 + \frac{R}{L}D + \frac{1}{LC}\right)i = 0
Auxiliary Equation will be:
m^2 + \frac{R}{L}m + \frac{1}{LC} = 0
m = \frac{-\frac{R}{L} \pm \sqrt{\left(\frac{R}{L}\right)^2 - 4 \times 1 \times \frac{1}{LC}}}{2 \times 1}
= \frac{-\frac{R}{L} \pm \sqrt{\frac{R^2}{L^2} - \frac{4}{LC}}}{2}
Given R^2C = 4L \implies R^2 = \frac{4}{LC} :
= \frac{-\frac{R}{L} \pm \sqrt{\frac{4L}{CL^2} - \frac{4}{LC}}}{2}
= \frac{-\frac{R}{L} \pm \sqrt{\frac{4}{LC} - \frac{4}{LC}}}{2}
\implies m = -\frac{R}{2L}, -\frac{R}{2L}
\text{C.F.} = (C_1 + C_2 t)e^{-\frac{R}{2L}t}
General solution:
i = (C_1 + C_2 t)e^{-\frac{R}{2L}t}
Example: 15. l \frac{d^2\theta}{dt^2} + g\theta = 0
Given that \theta = \theta_0 and \frac{d\theta}{dt} = 0 when t = 0.
Solution: l \frac{d^2\theta}{dt^2} + g\theta = 0
Given equation can be written as:
\left(D^2 + \frac{g}{l}\right)\theta = 0
Auxiliary equation will be:
m^2 + \frac{g}{l} = 0
m = \pm \sqrt{\frac{g}{l}}i
\text{C.F.} = C_1 \cos\left(\sqrt{\frac{g}{l}}t\right) + C_2 \sin\left(\sqrt{\frac{g}{l}}t\right)
General solution:
\theta = C_1 \cos\left(\sqrt{\frac{g}{l}}t\right) + C_2 \sin\left(\sqrt{\frac{g}{l}}t\right)
\frac{d\theta}{dt} = -C_1 \sqrt{\frac{g}{l}} \sin\left(\sqrt{\frac{g}{l}}t\right) + C_2 \sqrt{\frac{g}{l}} \cos\left(\sqrt{\frac{g}{l}}t\right)
\theta = \theta_0 when t = 0:
\theta_0 = C_1 \cos 0 + C_2 \sin 0 \implies C_1 = \theta_0
\frac{d\theta}{dt} = 0 when t = 0:
0 = -C_1 \sqrt{\frac{g}{l}} \sin 0 + C_2 \sqrt{\frac{g}{l}} \cos 0
\implies C_2 = 0
\theta = \theta_0 \cos\left(\sqrt{\frac{g}{l}}t\right)
By solving the above Examples, one can understand Complementary Function in BSc Maths.

Also Read This Article:- Simultaneous Equations B.Sc Maths

3.Practice Problems of Complementary Function in BSc Maths for Students

Solve following differential equations
(1.) \frac{d^2y}{dx^2} + a\frac{dy}{dx} + by = 0
(2.) \frac{d^4y}{dx^4} - a^4y = 0
Answers:
(2.) y = C_1 e^{ax} + C_2 e^{-ax} + C_3 \cos ax + C_4 \sin ax
By solving the above questions,you can understand the Complementary Function in BSc Maths well because the concept is well understood when you solve it practically.
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Also Read This Article:-  Simultaneous Differential Equation Bsc Maths

4.Frequently Asked Questions Related to Complementary Function in BSc Maths:

5.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

Q: 1. What is complementary function?

Ans: In general solution y=Y+u of the equation (D^n + a_1 D^{n-1} + \dots + a_n)y = X, then Y is called the complementary function (C.F.) and u is called the particular integral (P.I.).

Q: 2. What is called a particular integral?

Ans: The solution y of above equation can be determined by the methods discussed above. The problem is none to find the particular integral. We give certain methods of finding it.

Q: 3. What is meant by auxiliary equation?

Ans: Consider the differential equation
(D^n + a_1 D^{n-1} + a_2 D^{n-2} + \dots + a_n)y = 0
the equation becomes
(m^n + a_1 m^{n-1} + a_2 m^{n-2} + \dots + a_n)e^{mx} = 0
is called the Auxiliary equation.
By answering the above questions,you can know about the primary terms of Complementary Function in BSc Maths.

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*”निम्नलिखित प्रश्न में प्रश्नचिह्न (?) के स्थान पर क्या आएगा?”*
\frac{3 \times 16 + 20 \div 4}{6 \times 8 - 16 \div 2} = ?
(What will come in place of the question mark (?) in the following question?)
\frac{3 \times 16 + 20 \div 4}{6 \times 8 - 16 \div 2} = ?
*पिछली प्रश्नोत्तरी का हल*
माना पहली संख्या x तथा दूसरी संख्या y हो,तो प्रश्नानुसार
y + \frac{40x}{100} = y + \frac{60y}{100}
\implies 100y + 40x = 100y + 60y
\therefore \frac{x}{y} = \frac{60}{40} = \frac{3}{2} \implies x:y = 3:2
*Previous Quiz Solution*
Let the first number be x and the second number y, then according to the question
y + \frac{40x}{100} = y + \frac{60y}{100}
\implies 100y + 40x = 100y + 60y
\frac{x}{y} = \frac{60}{40} = \frac{3}{2} \implies x:y = 3:2
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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