1.Simultaneous Differential Equation Bsc Maths

Simultaneous Differential Equation Bsc Maths

In this article “Simultaneous Differential Equation Bsc Maths”,we will solve differential equations by elimination method and substitution method.In these differential equations will be a set of relations consisting of as may simultaneous equations.

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2.Simultaneous Differential Equation Bsc

Simultaneous Differential Equation Bsc Maths

  \text{ Example : 3. } \frac{d^2 x}{dt^2} - 3x - 4y = 0 ; \frac{d^2 y}{dt^2} + x + y = 0 \\ \text{ Solution :} \frac{d^2 x}{dt^2} - 3x - 4y = 0 ; \frac{d^2 y}{dt^2} + x + y = 0 \\ \text{ In symbolic manner the given} \\ \text{ equation can be written as} \\ (D^2 - 3)x - 4y = 0 \cdots (1) \\ x + (D^2 + 1)y = 0 \cdots (2) \\ \text{ Operation both sides of } (1) \text{ by} \\ (D^2 + 1) \text{ and multiplying } (2) \text{ by } 4, \text{ we obtain} \\ (D^2 + 1) \cdot (D^2 - 3)x - 4(D^2 + 1)y = 0 \cdots (3) \\ 4x + 4(D^2 + 1)y = 0 \cdots (4) \\ \text{ Adding } (3) \text{ and } (4), \text{ we obtain} \\ [(D^2 + 1)(D^2 - 3) + 4]x = 0 \\ \Rightarrow (D^4 - 2D^2 + 1)x = 0 \cdots (5)\\ \text{ which is linear equation with} \\ \text{ constant coefficients} \\ \text{ Auxiliary equation in this case} \text{ will be} \\ m^4 - 2m^2 + 1 = 0 \\ \Rightarrow (m^2 - 1)^2 = 0 \Rightarrow m = \pm 1, \pm 1\\ \text{ Now general solution of } (5) \text{ will} \text{ be} \\ x = (C_1 + C_2 t)e^t + (C_3 + C_4 t)e^{-t} \cdots (6) \\ \text{ Differentiation } (6) \text{ w.r.t. } t, \text{ we} \text{ obtain} \\ \frac{dx}{dt} = (C_1 + C_2 + C_2 t)e^t + (C_4 - C_3 - C_4 t)e^{-t} \\ \text{ Again differentiating} \\ \frac{d^2 x}{dt^2} = (C_1 + 2C_2 + C_2 t)e^t - (2C_4 - C_3 - C_4 t)e^{-t} \\ \text{ Substituting this value of } \frac{d^2 x}{dt^2} \\ \text{ and the value of } x \text{ from equation} (6) \text{ in } (1), \text{ we obtain} \\ 4y = D^2 x - 3x \\ = (C_1 + 2C_2 + C_2 t)e^t - (2C_4 - C_3 - C_4 t)e^{-t} - 3 [(C_1 + C_2 t)e^t + (C_3 + C_4 t)e^{-t}] \\ \therefore y = \frac{1}{2}(C_2 - C_1 - C_2 t)e^t - \frac{1}{2}(C_4 + C_3 + C_4 t)e^{-t} \\ \text{ and } x = (C_1 + C_2 t)e^t + (C_3 + C_4 t)e^{-t} 
\text{ Example : 4. } \frac{d^2 x}{dt^2} + m^2 y = 0 ; \frac{d^2 y}{dt^2} - m^2 x = 0 \\ \text{ Solution :} \frac{d^2 x}{dt^2} + m^2 y = 0 ; \frac{d^2 y}{dt^2} - m^2 x = 0 \\ \text{ The given equation can be written} \text{ as} \\ D^2 x + m^2 y = 0 \cdots (1) \\ -m^2 x + D^2 y = 0 \cdots (2) \\ \text{ Operation } (1) \text{ by } D^2 \text{ and multiplying} \\ (2) \text{ by } m^2 \text{ and then subtracting} \\ \text{ the results so obtained, we get} \\ (D^4 + m^4)x = 0 \cdots (3) \\ \text{ Its auxiliary equation is} \\ M^4 + m^4 = 0 \\ \Rightarrow (M^4 + 2m^2 M^2 + m^4) - 2m^2 M^2 = 0 \\ \Rightarrow (M^2 + m^2)^2 - (m\sqrt{2}M)^2 = 0\\ \Rightarrow (M^2 + \sqrt{2}mM + m^2)(M^2 - \sqrt{2}mM + m^2) = 0 \\ \text{ From } M^2 - \sqrt{2}mM + m^2 = 0, \text{ we get} \\ M = \frac{1}{2}[\sqrt{2}m \pm \sqrt{(2m^2 - 4m^2)}] = \frac{m \pm im}{\sqrt{2}} \\ \text{ Similarly from } M^2 + \sqrt{2}mM + m^2 = 0, \text{ we get} \\ M = \frac{-m \pm im}{\sqrt{2}} \\ \therefore \text{ The solution of } (3) \text{ is} \\ x = \left[C_1 \cos\left(\frac{mt}{\sqrt{2}}\right) + C_2 \sin\left(\frac{mt}{\sqrt{2}}\right)\right] e^{\frac{mt}{\sqrt{2}}} + \left[C_3 \cos\left(\frac{mt}{\sqrt{2}}\right) + C_4 \sin\left(\frac{mt}{\sqrt{2}}\right)\right] e^{-\frac{mt}{\sqrt{2}}} \cdots (4) \\ \frac{dx}{dt} = \frac{m}{\sqrt{2}} \left[C_1 \cos\left(\frac{mt}{\sqrt{2}}\right) + C_2 \sin\left(\frac{mt}{\sqrt{2}}\right)\right] e^{\frac{mt}{\sqrt{2}}} + \left[-\frac{C_1 m}{\sqrt{2}} \sin\left(\frac{mt}{\sqrt{2}}\right) + \frac{C_2 m}{\sqrt{2}} \cos\left(\frac{mt}{\sqrt{2}}\right)\right] e^{\frac{mt}{\sqrt{2}}} + \left[-\frac{C_3 m}{\sqrt{2}} \sin\left(\frac{mt}{\sqrt{2}}\right) + \frac{C_4 m}{\sqrt{2}} \cos\left(\frac{mt}{\sqrt{2}}\right)\right] e^{-\frac{mt}{\sqrt{2}}} - \frac{m}{\sqrt{2}} \left[C_3 \cos\left(\frac{mt}{\sqrt{2}}\right) + C_4 \sin\left(\frac{mt}{\sqrt{2}}\right)\right] e^{-\frac{mt}{\sqrt{2}}} \\ \frac{dx}{dt} = \frac{m}{\sqrt{2}} \left[(C_1 + C_2) \cos\left(\frac{mt}{\sqrt{2}}\right) + (C_2 - C_1) \sin\left(\frac{mt}{\sqrt{2}}\right)\right] e^{\frac{mt}{\sqrt{2}}} - \frac{m}{\sqrt{2}} \left[(C_3 + C_4) \sin\left(\frac{mt}{\sqrt{2}}\right) + (C_3 - C_4)\right] \cos\left(\frac{mt}{\sqrt{2}}\right) e^{-\frac{mt}{\sqrt{2}}} \\ \therefore D^2 x = \frac{1}{2}m \left[(C_1 + C_2) \cos\left(\frac{mt}{\sqrt{2}}\right) + (C_2 - C_1)\right] \sin\left(\frac{mt}{\sqrt{2}}\right) e^{\frac{mt}{\sqrt{2}}} + \frac{m}{\sqrt{2}} \left[-\frac{m}{\sqrt{2}} (C_1 + C_2)\right] \sin\left(\frac{mt}{\sqrt{2}}\right) - (C_2 - C_1)\frac{m}{\sqrt{2}} \cos\left(\frac{mt}{\sqrt{2}}\right) e^{\frac{mt}{\sqrt{2}}} + \frac{1}{2}m \left[(C_3 + C_4) \sin\left(\frac{mt}{\sqrt{2}}\right) + (C_3 - C_4) \cos\left(\frac{mt}{\sqrt{2}}\right)\right] e^{-\frac{mt}{\sqrt{2}}} - \frac{m}{\sqrt{2}} \left[\frac{m}{\sqrt{2}} (C_3 + C_4) \cos\left(\frac{mt}{\sqrt{2}}\right) - (C_3 - C_4)\right] \frac{m}{\sqrt{2}} \sin\left(\frac{mt}{\sqrt{2}}\right) e^{-\frac{mt}{\sqrt{2}}} \\ \Rightarrow \frac{d^2 x}{dt^2} = m \left[\left(C_2 \cos\left(\frac{mt}{\sqrt{2}}\right) - C_1 \sin\left(\frac{mt}{\sqrt{2}}\right)\right)\right] e^{\frac{mt}{\sqrt{2}}} + \left[\left(C_3 \sin\left(\frac{mt}{\sqrt{2}}\right) - C_4 \cos\left(\frac{mt}{\sqrt{2}}\right)\right)\right] e^{-\frac{mt}{\sqrt{2}}} \cdots (5) \\ \text{ From } (1), \text{ we get} \\ m^2 y = -D^2 x \\ \Rightarrow y = -\frac{1}{m^2} (D^2 x) \\ \Rightarrow y = -\frac{1}{m^2} \left[\left(C_2 \cos\frac{mt}{\sqrt{2}} - C_1 \sin\frac{mt}{\sqrt{2}}\right) e^{\frac{mt}{\sqrt{2}}}+ \left(C_3 \sin\frac{mt}{\sqrt{2}} - C_4 \cos\frac{mt}{\sqrt{2}}\right) e^{-\frac{mt}{\sqrt{2}}}\right]
\text{ Example : 5. } \frac{dx}{dt} + 2\frac{dy}{dt} - 2x + 2y = 3e^t , 3\frac{dx}{dt} + \frac{dy}{dt} + 2x + y = 4e^{2t} \\ \text{ Solution : } \frac{dx}{dt} + 2\frac{dy}{dt} - 2x + 2y = 3e^t , 3\frac{dx}{dt} + \frac{dy}{dt} + 2x + y = 4e^{2t} \\ \text{ The given equation can be rewritten} \text{ as} \\ (D-2)x + (2D+2)y = 3e^t \cdots (1) \\ (3D+2)x + (D+1)y = 4e^{2t} \cdots (2) \\ \text{ multiplying } (2) \text{ by } 2 \text{ and subtracting} \\ \text{ from it } (1), \text{ we get} \\ \begin{array}{c}(D-2)x + (2D+2)y = 3e^t \cdots (1) \\ 2(3D+2)x + (2D+2)y = 8e^{2t} \cdots (3) \\ \hline \end{array} \\ (D - 2 - 6D - 4)x = 3e^t - 8e^{2t} \\ \Rightarrow (-5D - 6)x = 3e^t - 8e^{2t} \\ (5D + 6)x = 8e^{2t} - 3e^t \\ \text{ Its auxiliary equation is} \\ 5m + 6 = 0 \Rightarrow m = -\frac{6}{5} \\ \text{ C.F.} = C_1 e^{-\frac{6}{5}t} \\ \frac{dx}{dt} + \frac{6}{5}x = \frac{8}{5}e^{2t} - \frac{3}{5}e^t\\ \text{ It is Linear Differential Equation} \\ \text{ Integrating factor} = e^{\int P dt} \\ = e^{\int \frac{6}{5} dt} = e^{\frac{6}{5}t} \\ \Rightarrow x \cdot \text{ I.F.} = \int (\text{ I.F.}) Q dt + C_1 \\ \therefore x e^{\frac{6}{5}t} = \int \left(\frac{8}{5}e^{2t} - \frac{3}{5}e^t\right) e^{\frac{6}{5}t} dt + C_1 \\ \Rightarrow x e^{\frac{6}{5}t} = \int \left(\frac{8}{5}e^{\frac{16}{5}t} - \frac{3}{5}e^{\frac{11}{5}t}\right) dt + C_1 \\ = \frac{8}{5} \times \frac{5}{16} e^{\frac{16}{5}t} - \frac{3}{5} \times \frac{5}{11} e^{\frac{11}{5}t} + C_1 \\ \Rightarrow x = \frac{\frac{1}{2} e^{\frac{16}{5}t}}{e^{\frac{6}{5}t}} - \frac{\frac{3}{11} e^{\frac{11}{5}t}}{e^{\frac{6}{5}t}} + C_1 e^{-\frac{6}{5}t} \\ \Rightarrow x = \frac{1}{2} e^{2t} - \frac{3}{11} e^t + C_1 e^{-\frac{6}{5}t} \cdots (4) \\ \text{ Differentiating w.r.t. } t, \text{ we obtain} \\ \frac{dx}{dt} = e^{2t} - \frac{3}{11} e^t - \frac{6}{5} C_1 e^{-\frac{6}{5}t} \\ \text{ substituting this value of } \frac{dx}{dt} \\ \text{ and value of } x \text{ from equation} \\ (4) \text{ in } (1), \text{ we obtain} \\ \left(e^{2t} - \frac{3}{11}e^t - \frac{6}{5}C_1 e^{-\frac{6}{5}t}\right) + 2\frac{dy}{dt} - 2\left(\frac{1}{2} e^{2t} - \frac{3}{11}e^t\right. \left.+ C_1 e^{-\frac{6}{5}t}\right) + 2y = 3e^t \\ \Rightarrow e^{2t} - \frac{3}{11}e^t - \frac{6}{5}C_1 e^{-\frac{6}{5}t} + 2\frac{dy}{dt} - e^{2t} + \frac{6}{11}e^t - 2C_1 e^{-\frac{6}{5}t} + 2y = 3e^t \\ \Rightarrow \frac{3}{11}e^t - \frac{16}{5}C_1 e^{-\frac{6}{5}t} + 2\frac{dy}{dt} + 2y = 3e^t \\ \Rightarrow 2\frac{dy}{dt} + 2y = \frac{30}{11}e^t + \frac{16}{5}C_1 e^{-\frac{6}{5}t} \\ \Rightarrow \frac{dy}{dt} + y = \frac{15}{11}e^t + \frac{8}{5}C_1 e^{-\frac{6}{5}t} \\ \text{ I.F.} = e^{\int P dt} = e^{\int 1 dt} = e^t \\ y \cdot e^t = \int \left(\frac{15}{11}e^t + \frac{8}{5}C_1 e^{-\frac{6}{5}t}\right) e^t dt + C_2 \\ = \int \left(\frac{15}{11}e^{2t} + \frac{8}{5}C_1 e^{-\frac{1}{5}t}\right) dt + C_2 \\ = \frac{15}{11} \times \frac{1}{2} e^{2t} + C_1 \frac{8}{5} \times -5 e^{-\frac{1}{5}t} + C_2\\ \Rightarrow y e^t = \frac{15}{22} e^{2t} - 8C_1 e^{-\frac{1}{5}t} + C_2 \\ \Rightarrow y = \frac{15}{22} e^{2t - t} - 8C_1 e^{-\frac{1}{5}t - t} + C_2 e^{-t} \\ \Rightarrow y = \frac{15}{22} e^t - 8C_1 e^{-\frac{6}{5}t} + C_2 e^{-t} \\ \text{ Example : 6. } t\frac{dx}{dt} + y = 0 ; \ t\frac{dy}{dt} + x = 0 \\ \text{ Solution : } t\frac{dx}{dt} + y = 0 \cdots (1) ; \ t\frac{dy}{dt} + x = 0 \cdots (2) \\ \text{ Differentiating } (1) \text{ w.r.t. } t, \text{ we obtain} \\ t\frac{d^2 x}{dt^2} + \frac{dx}{dt} + \frac{dy}{dt} = 0 \cdots (3) \\ \text{ Substituting this value of } \frac{dy}{dt} \text{ from} \\ (3) \text{ in } (2), \text{ we obtain} \\ t \left(t\frac{d^2 x}{dt^2} + \frac{dx}{dt} + \frac{dy}{dt}\right) + x = 0 \\ \Rightarrow t^2 \frac{d^2 x}{dt^2} + t\frac{dx}{dt} + x = 0 \\ \Rightarrow t^2 \frac{d^2 x}{dt^2} + t\frac{dx}{dt} - x = 0 \cdots (4) \\ \text{ This is homogeneous linear} \text{ equation of 2nd order} \\ \therefore \text{ Put } z = \log_e t \text{ in } (4) \Rightarrow t = e^z \\ \text{ Now equation becomes like} \text{ this} \\ [D(D-1) + (D-1)]x = 0, \text{ where } D = \frac{d}{dz} \\ (D^2 - 1)x = 0 \cdots (5) \\ \text{ Its auxiliary equation is} \\ m^2 - 1 = 0 \Rightarrow m = \pm 1 \\ \therefore x = C_1 e^z + C_2 e^{-z} \\ \Rightarrow x = C_1 t + C_2 t^{-1} \cdots (6) \\ \text{ Again from } (6), \ \frac{dx}{dt} = C_1 - C_2 t^{-2} \\ \text{ Now putting this value in } (1),\\ t(C_1 - C_2 t^{-2}) + y = 0 \\ \therefore y = -C_1 t + C_2 t^{-1} \cdots (7) \\ \text{ and } x = C_1 t + C_2 t^{-1} 

By solving the above questions,one can understand the Simultaneous Differential Equation Bsc Maths well because the concept is well understood when you solve the questions practically.

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3.Practice Questions of Simultaneous Differential Equation Bsc Maths for Students

Simultaneous Differential Equation Bsc Maths

Solve the following differential equations
(1.)  4\frac{dx}{dt} + 9\frac{dy}{dt} + 2x + 31t = e^t,3\frac{dx}{dt} + 7\frac{dy}{dt} + x + 24y = 3 \\ (2.)  2\frac{d^2 y}{dx^2} - \frac{dz}{dx} - 4y = 2x , 2\frac{dy}{dx} + 4\frac{dz}{dx} - 3z = 0 \\ \text{Answers : } (1.) x = \frac{1}{2} [-(C_1 + C_2)\sin t + (C_1 - C_2)\cos t] e^{-4t} + \frac{27}{26} e^t - \frac{23}{17} \\ y = e^{-4t} \{C_1 \sin t + C_2 \cos t\} - \frac{2}{13} e^t + \frac{6}{17} \\ (2.) y = (C_1 + C_2 x)e^x + C_3 e^{-\frac{3}{2}x} - \frac{1}{2}x \\ z = -2e^x (C_1 + C_2 x - 3C_2) - \frac{1}{3} C_3 e^{-\frac{3}{2}x} - \frac{1}{3}
By solving the above questions,one can understand the Simultaneous Differential Equation Bsc Maths well because the concept is well understood when you solve the questions practically.

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4.Frequently Asked Questions Related to Simultaneous Differential Equation Bsc Maths

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

Simultaneous Differential Equation 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!👇
*”ABC एक त्रिभुज है,जिसका आधार AB है।AB पर D एक ऐसा बिन्दु है कि AB=5 और DB=3 तो \triangle ADC के क्षेत्रफल का \triangle  ABC के क्षेत्रफल से अनुपात क्या होगा?”*
(ABC is a triangle, the basis of which is AB. D on AB is a point that AB=5 and DB=3 So what is the ratio of the area of \triangle ADC to the area of \triangle ABC?)
*पिछली प्रश्नोत्तरी का हल*
परिमेय करण करने पर :
\frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}} \\ = \frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} \\ = \frac{\sqrt{5} + \sqrt{10} - \sqrt{3} - \sqrt{6}}{5 - 3} + \frac{\sqrt{5} - \sqrt{10} + \sqrt{3} - \sqrt{6}}{5 - 3} \\ = \frac{1}{2}(2\sqrt{5} - 2\sqrt{6}) = \sqrt{5} - \sqrt{6}
*Previous Quiz Solution*
Rationalization :
\frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}} \\ = \frac{1 + \sqrt{2}}{\sqrt{5} + \sqrt{3}} \times \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{1 - \sqrt{2}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} \\ = \frac{\sqrt{5} + \sqrt{10} - \sqrt{3} - \sqrt{6}}{5 - 3} + \frac{\sqrt{5} - \sqrt{10} + \sqrt{3} - \sqrt{6}}{5 - 3} \\ = \frac{1}{2}(2\sqrt{5} - 2\sqrt{6}) = \sqrt{5} - \sqrt{6}
This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*