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How to Solve Simultaneous Equations BSc Maths

September 25, 2026 Differential Equation No Comments
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1 1.How to Solve Simultaneous Equations BSc Maths
1.1 2.Key Points of Simultaneous Equations BSc Maths
1.2 3.Examples Based on Simultaneous Equations BSc Maths
1.2.1 4.Practice Problems of Simultaneous Equations BSc Maths for Students
1.2.2 5.Frequentaly Asked Questions Related to Simultaneous Equations BSc Maths
1.2.3 Q:1.What is the difference between partial and ordinary differential equations?
1.2.4 Q:2.Does the degenerate system have a solution,Clarify?
1.2.5 Q:3.What is the condition for Simultaneous Equations BSc Maths to be generated?
1.3 6.छात्र-छात्राओं से आज का प्रश्न (Today’s Question to Students):

1.How to Solve Simultaneous Equations BSc Maths

Do you face difficulty to Solve Simultaneous Equations BSc Maths. We will help you by simple method and complete guide of above topic.
“Before moving to Part 2,read our basic introduction to Simultaneous Differential Equation Bsc.”

Also Read This Article:- Simultaneous Differential Equation Bsc Maths

2.Key Points of Simultaneous Equations BSc Maths

(1) Definition: Simultaneous linear ordinary differential equations involve more than two variables: one independent variable and multiple dependent variables (x, y, \dots).
(2) Standard Form: Represented symbolically as f_1(D)x + f_2(D)y = \phi_1(t), g_1(D)x + g_2(D)y = \phi_2(t)
Where D = \frac{d}{dt} and f_1, f_2, g_1, g_2 are polynomial differential operators.
(3)Solving Method (Elimination): Eliminate one dependent variable (e.g., y) using algebraic operator multiplication.
(4)Solve the single-variable linear differential equation with constant coefficients for x.
(5)Substitute x back or repeat the elimination process to find y.
(6.) Determinant Representation:
\begin{vmatrix} f_1(D) & f_2(D) \\ g_1(D) & g_2(D) \end{vmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{vmatrix} \phi_1(t) & f_2(D) \\ \phi_2(t) & g_2(D)\end{vmatrix}
(7.)Degeneracy Condition: System is solvable only if the operator determinant \Delta(D) \neq 0
(8.)If \Delta(D)= 0, the system is degenerate and may have either no solution or infinitely many solutions.

Also Read This Article:- Singular Solutions,Envelope and Cusp Locus-B.Sc. Maths

3.Examples Based on Simultaneous Equations BSc Maths

Example:7 t \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + tx = 0
Solution: t \frac{d^2x}{dt^2} + 2 \frac{dx}{dt} + tx = 0 \quad \dots (1)
Put tx = v \quad \dots (3)
Differentiating with respect to t, we get
x + t\frac{dx}{dt} = \frac{dv}{dt}
Again differentiating with respect to t, we get
\frac{dx}{dt} + \frac{dx}{dt} + t\frac{d^2x}{dt^2} = \frac{d^2v}{dt^2}
\Rightarrow t\frac{d^2x}{dt^2} + 2\frac{dx}{dt} = \frac{d^2v}{dt^2} \quad \dots (4)
From Equation (1), (3) and (4), we get
\frac{d^2v}{dt^2} + v = 0
\Rightarrow (D^2 + 1)v = 0
Auxiliary equation for above equation will be
m^2 + 1 = 0 \implies m = \pm i
General solution of this equation will be
v = c_1 \cos t + c_2 \sin t
Since tx = v \implies x = \frac{v}{t}
x = \frac{c_1 \cos t + c_2 \sin t}{t}
v = c_1 \cos t + c_2 \sin t
\Rightarrow tx = c_1 \cos t + c_2 \sin t
\Rightarrow x = \frac{c_1}{t} \cos t + \frac{c_2}{t} \sin t \quad \dots (5)
Differentiating with respect to t of above equation
\frac{dx}{dt} = -\frac{c_1}{t^2}\cos t - \frac{c_1}{t}\sin t - \frac{c_2}{t^2}\sin t + \frac{c_2}{t}\cos t \quad \dots (6)
Putting the value of \frac{dx}{dt} in equation (2), we get
\frac{dy}{dt} + \frac{2}{t}y = -\frac{c_1}{t^2}\cos t - \frac{c_1}{t}\sin t - \frac{c_2}{t^2}\sin t + \frac{c_2}{t}\cos t
\Rightarrow t^2\frac{dy}{dt} + 2ty = -c_1\cos t - c_1 t\sin t - c_2\sin t + c_2 t\cos t \quad \dots (7)
Put t^2 y = p \quad \dots (8)
Differentiation equation (8) with respect to t, we get
t^2\frac{dy}{dt} + 2ty = \frac{dp}{dt} \quad \dots (9)
From equation (7) and (9), we get
\frac{dp}{dt} = -c_1\cos t - c_1 t\sin t - c_2\sin t + c_2 t\cos t
\Rightarrow \int dp = \int (-c_1\cos t)dt - c_1 \int t\sin t,dt - c_2 \int \sin t dt + c_2 \int t\cos t dt
\Rightarrow p = -c_1\sin t - c_1 \int t\sin t dt + c_2\cos t + c_2 \int t \cos t dt
Example: 8. lt \frac{dx}{dt} = mn(y-z); mt \frac{dy}{dt} = nl(z-x); nt \frac{dz}{dt} = lm(x-y)
Solution: lt \frac{dx}{dt} = mn(y-z) \quad \cdots{ (1)}
mt \frac{dy}{dt} = nl(z-x) \quad \cdots{ (2)}
nt \frac{dz}{dt} = lm(x-y) \quad \cdots{ (3)}
Put lx = X, my = Y, nz = Z, t = e^T \quad \cdots{ (4)}
Differentiating with respect to t:
l \frac{dx}{dt} = \frac{dX}{dt} = \frac{dT}{dt} \cdots \frac{dX}{dT} = \frac{1}{e^T} \frac{dX}{dT} \quad \cdots{ (5)}
m \frac{dy}{dt} = \frac{dY}{dt} = \frac{dT}{dt} \cdots \frac{dY}{dT} = \frac{1}{e^T} \frac{dY}{dT} \quad \cdots{ (6)}
n \frac{dz}{dt} = \frac{dZ}{dt} = \frac{dT}{dt} \cdots \frac{dZ}{dT} = \frac{1}{e^T} \frac{dZ}{dT} \quad \cdots{ (7)}
\frac{dt}{dT} = e^T \quad \cdots{ (8)}
From (1), (4) and (5), we get:
t \cdots \left(\frac{1}{e^T}\right) \frac{dX}{dT} = ny-mz \implies \frac{dX}{dT} = nY - mZ [\because e^T=t] \quad \cdots{ (9)}
From (2), (4) and (6), we get:
t \cdot \left(\frac{1}{e^T}\right) \frac{dY}{dT} = lZ - nX \\ \implies \frac{dY}{dT}=lz-nX [\because e^T=t] \cdots{ (10)}
From (3), (4) and (7), we get:
e^T \cdot \left(\frac{1}{e^T} \frac{dZ}{dT}\right) = mX - lY \implies \frac{dZ}{dT} = mX - lY \quad \cdots{ (11)}
Multiplying (9), (10), (11) by 2X, 2Y and 2Z respectively
2X \frac{dX}{dT} + 2Y \frac{dY}{dT} + 2Z \frac{dZ}{dT} = 0
\implies \frac{d}{dT}(X^2 + Y^2 + Z^2) = 0
Integrating
X^2 + Y^2 + Z^2 = c_1 \quad \cdots{ (12)}
Now Multiplying (9), (10) and (11) by l, m and n respectively
l \frac{dX}{dT} + m \frac{dY}{dT} + n \frac{dZ}{dT} = 0
\frac{d}{dT}(lX + mY + nZ) = 0
Integrating
\implies lX + mY + nZ = c_2 \quad \cdots{ (13)}
Differentiating (9), (10) and (11) with respect to T, we get
\frac{d^2X}{dT^2} = n \frac{dY}{dT} - m \frac{dZ}{dT} \quad \cdots{ (14)}
\frac{d^2Y}{dT^2} = l \frac{dZ}{dT} - n \frac{dX}{dT} \quad \cdots{ (15)}
\frac{d^2Z}{dT^2} = m \frac{dX}{dT} - l \frac{dY}{dT} \quad \cdots{ (16)}
Now multiplying (14), (15) and (16) by 2\frac{dX}{dT}, 2\frac{dY}{dT}, 2\frac{dZ}{dT} respectively and adding
2\frac{dX}{dT} \frac{d^2X}{dT^2} + 2\frac{dY}{dT} \frac{d^2Y}{dT^2} + 2\frac{dZ}{dT} \frac{d^2Z}{dT^2} = 0
Integrating:
\left(\frac{dX}{dT}\right)^2 + \left(\frac{dY}{dT}\right)^2 + \left(\frac{dZ}{dT}\right)^2 = c_3
(nY - mZ)^2 + (lZ - nX)^2 + (mX - lY)^2 = c_3 \quad \cdots{ (17)}
Equation (12), (13) and (17) gives us general solution where lX = x, mY = y, nZ = z, t = e^T

Example: 9. Solve the equations: \frac{dx}{dt} = -\omega y and \frac{dy}{dt} = \omega x and show that the point (x, y) lies on a circle.
Solution: Given equation is
\frac{dx}{dt} = -\omega y \quad \cdots{ (1)}
\frac{dy}{dt} = \omega x \quad \cdots{ (2)}
Differentiating (1) with respect to t, we get
\frac{d^2x}{dt^2} = -\omega \frac{dy}{dt} = -\omega^2 x \quad [\text{From (2)}]
\implies (D^2 + \omega^2)x = 0 \quad \cdots{ (3)}
Auxiliary equation will be
m^2 + \omega^2 = 0 \implies m = \pm i\omega
\therefore x = c_1 \cos\omega t + c_2 \sin\omega t \quad \cdots{ (4)}
Again from (4)
\frac{dx}{dt} = -c_1 \omega \sin\omega t + c_2 \omega \cos\omega t
Putting value of \frac{dx}{dt} in (1), we get
-c_1 \omega \sin\omega t + c_2 \omega \cos\omega t = -\omega y
\therefore y = c_1 \sin\omega t - c_2 \cos\omega t \quad \cdots{ (5)}
Now adding squaring of (4) and (5), we get
x^2 + y^2 = c_1^2 + c_2^2 \quad \cdots{ (6)}
Which is equation of circle.
Therefore (x, y) lies on circle (6).
Example: 10. Prove that the general equation solution of the simultaneous equations
m \frac{d^2x}{dt^2} = Ve - He \frac{dy}{dt}, m \frac{d^2y}{dt^2} = He \frac{dx}{dt}
where m, V, H, e are constants, is
x = a + b \cos(\omega t - \alpha), y = \frac{Vt}{H} + c + b \sin(\omega t - \alpha)
where \omega = \frac{He}{m} and a, b, c, \alpha are arbitrary constants.
Solution:
m \frac{d^2x}{dt^2} = Ve - He \frac{dy}{dt}
m \frac{d^2y}{dt^2} = He \frac{dx}{dt}
Given Equation can be written as
mD^2 x + HeDy = Ve \quad \cdots{ (1)}
HeDx - mD^2 y = 0 \quad \cdots{ (2)}
Multiplying (1) by mD and (2) by He, we get
m^2 D^3 x + mHe D^2 y = D(Ve)
\implies m^2 D^3 x + mHe D^2 y = 0 \quad \cdots{ (3)}
H^2 e^2 Dx - mHe D^2 y = 0 \quad \cdots{ (4)}
Adding (3) and (4)
m^2 D^3 x + H^2 e^2 Dx = 0
\implies (m^2 D^3 + H^2 e^2 D)x = 0
Now Auxiliary equation will be
m^2 M^3 + H^2 e^2 M = 0
M (m^2 M^2 + H^2 e^2) = 0
\implies M = 0, \pm \frac{He}{m}i
x = a + c_1 \cos\left(\frac{He}{m}t\right) + c_2 \sin\left(\frac{He}{m}t\right)
\text{Put } c_1 = b \cos\alpha, c_2 = b \sin\alpha, \frac{He}{m} = \omega
x = a + b \cos\alpha \cos\omega t + b \sin\alpha \sin\omega t
x = a + b \cos(\omega t - \alpha)
Differentiating with respect to t, we get
\frac{dx}{dt} = -b\omega \sin(\omega t - \alpha)
Again differentiating with respect to t, we get
\frac{d^2x}{dt^2} = -b\omega^2 \cos(\omega t - \alpha)
Put above value in equation (1)
m \left[-b\omega^2 \cos(\omega t - \alpha)\right] = Ve - He \frac{dy}{dt}
\implies He \frac{dy}{dt} = Ve + mb\omega^2 \cos(\omega t - \alpha)
\implies \frac{dy}{dt} = \frac{Ve}{He} + \frac{mb\omega^2}{He} \cos(\omega t - \alpha)
\implies \frac{dy}{dt} = \frac{V}{H} + \frac{1}{\omega} b\omega^2 \cos(\omega t - \alpha) \quad \left[\because \frac{m}{He} = \frac{1}{\omega}\right]
\implies \frac{dy}{dt} = \frac{V}{H} + b\omega \cos(\omega t - \alpha)
Integrating:
\int dy = \int \frac{V}{H} dt + b\omega \int \cos(\omega t - \alpha) dt
\implies y = \frac{Vt}{H} + b \sin(\omega t - \alpha) + c
and x = a + b \cos(\omega t - \alpha)
By solving the above examples,one can understand the Simultaneous Equations BSc Maths.

Also Read This Article:- Simultaneous equations

4.Practice Problems of Simultaneous Equations BSc Maths for Students

(1) Solve \frac{d^2x}{dt^2} + 4x + y = t^2 e^t,
\frac{d^2y}{dt^2} + y - 2x = \cos^2 t
(2) Solve 4 \frac{dx}{dt} + 9 \frac{dy}{dt} + 44x + 49y = 0 ,3 \frac{dx}{dt} + 7 \frac{dy}{dt} + 34x + 38y = e^t
Answers: (1.) x = (c_1 \cos\sqrt{3}t + c_2 \sin\sqrt{3}t) + (c_3 \cos\sqrt{2}t + c_4 \sin\sqrt{2}t) - \frac{42}{1452} e^{3t} + \frac{57}{66} e^{3t} + \frac{1}{12} - \frac{1}{4} \cos 2t , \\ y=- 3c_2 \sin\sqrt{3}t) - c_2 \sin\sqrt{3}t - 2c_3 \cos\sqrt{2}t - 2c_4 \sin\sqrt{2}t + \frac{1}{66} e^{3t} + \frac{1}{3} - \frac{23}{1452} e^{3t}
(2.) x = c_1 e^{-t} + c_2 e^{-6t} + \frac{19}{3} t - \frac{56}{9} - \frac{29}{7} e^t , \\ y = c_1 e^{-t} + 4c_2 e^{-6t} + \frac{17}{3} t + \frac{55}{9} + \frac{24}{7} e^t
By solving the above questions,you can understand the Simultaneous Equations BSc Maths well because the concept is well understood when you solve it practically.

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5.Frequentaly Asked Questions Related to Simultaneous Equations BSc Maths

Q:1.What is the difference between partial and ordinary differential equations?

Ans:The ordinary equation will have only one independent variable while the partial differential equations will have more than one.

Q:2.Does the degenerate system have a solution,Clarify?

Ans:Degenerate may have either no solution or infinitely many solutions.

Q:3.What is the condition for Simultaneous Equations BSc Maths to be generated?

Ans: \begin{array}{c} f_1(D) & f_2(D) \\ g_1(D) & g_2(D) \end{array} = 0, the system of simultaneous equation is said to be degenerate.
The system of Simultaneous Differential Equation is said to be degenerate.
By answering the above questions,you can know about the primary terms of Simultaneous Equations BSc Maths.

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

🎯 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!👇
*”दो व्यक्ति एक चौराहे से दो सड़कों पर चलते हैं जो 60° का कोण बनाती हैं।वे 3 किमी/घण्टे की एकसी चाल से चलते हैं।20 मिनट बीतने पर वे एक-दूसरे से (कम से कम) कितनी दूरी पर होंगे?”*
(Two persons walk from an intersection on two roads that form an angle of 60°. They walk at the same speed of 3 km/hr. How far will they be (at least) from each other after the passage of 20 minutes?)
(What is the adjacent value of of 331?)
*पिछली प्रश्नोत्तरी का उत्तर*
25% के लाभ पर साबुन के घोल का मूल्य प्रति लीटर में=16 \times \frac{125}{100} 
=20 रु. प्रति लीटर
तब 20 रु. में बेचा गया द्रव=\frac{20}{18} लीटर
अतः अभीष्ट अनुपात=]1 : \left(\frac{20}{18} – 1\right) = 1 : \frac{2}{18} = 9 : 1
*Previous Quiz Solution*
The cost of soap solution at a profit of 25% in per litre =16 \times \frac{125}{100} \\ =20    Rs. per litre
Then the liquid sold for Rs. 20=\frac{20}{18}  litre
Therefore,the desired ratio=]1 : \left(\frac{20}{18} – 1\right) = 1 : \frac{2}{18} = 9 : 1
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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