Differential Equations
What are Differential Equations? – Introduction
Differential equations are mathematical equations that involve derivatives of an unknown function. They are used to model various real-world phenomena and describe how a quantity changes over time or space. Unlike ordinary equations, which involve only algebraic operations, differential equations bring in the concept of differentiation to express the relationship between the unknown function and its derivatives.
Types of Differential Equations
Ordinary Differential Equations (ODEs): These differential equations involve a single independent variable and its derivatives with respect to that variable. ODEs are often used to describe dynamic systems such as population growth, radioactive decay, and electrical circuits.
Partial Differential Equations (PDEs): These differential equations involve multiple independent variables and their partial derivatives. PDEs are commonly used to describe physical phenomena like heat transfer, fluid dynamics, and electromagnetic fields.
Solving Differential Equations
Solving differential equations requires finding the unknown function that satisfies the given equation. The approach to solving differential equations depends on their type and order. Various methods are employed, including separation of variables, integrating factors, substitution, and using special functions such as exponential, logarithmic, and trigonometric functions.
Applications of Differential Equations
Physics: Differential equations are extensively used in physics to model and understand phenomena such as motion, wave propagation, quantum mechanics, and electromagnetism.
Engineering: Differential equations play a crucial role in engineering disciplines like mechanical, civil, electrical, and chemical engineering. They are used to analyse systems such as control systems, fluid flow, structural mechanics, and circuit behaviour.
Biology: Differential equations are employed to study biological processes, including population dynamics, enzyme kinetics, neural networks, and epidemiology.
Economics: Differential equations find applications in economic modelling, particularly in areas such as growth models, optimization, and game theory.
Differential Equations – Exam Questions
Q1.
Solution
\begin{gathered} \frac{d y}{d x}=\frac{2-x}{y-3} \\ \int(y-3) d y=\int(2-x) d x \\ \frac{y^2}{2}-3 y=2 x-\frac{x^2}{2}+c \end{gathered}\begin{gathered} y=4 \quad x=5: \\ \frac{16}{2}-12=2(5)-\frac{5^2}{2}+c \\ \therefore c=-\frac{3}{2} \end{gathered}\Rightarrow \frac{y^2}{2}-3 y+\frac{x^2}{2}-2 x+\frac{3}{2}=0Part ii)
\begin{aligned} & y^2-6 y+x^2-4 x+3=0 \\ & (y-3)^2+(x-2)^2+3-9-4=0 \\ & (y-3)^2+(x-2)^2=10 \\ & a=2 \quad b=3 \quad k=10 \end{aligned}Part iii)
Circle centre (2, 3) radius \sqrt{10}
Question 2
Solution
\begin{aligned} & \frac{d h}{d t}=\frac{6-h}{20} \\ & -\int \frac{-1}{6-h} d h=\frac{1}{20} \int d t \\ & -\ln (6-h) \quad=\frac{1}{20} t+c \\ & t=0 \quad h=1 \quad \therefore \quad c=-\ln 5 \\ & \Rightarrow \quad-\ln (6-h)=\frac{1}{20} t-\ln 5 \\ & \quad \ln 5-\ln (6-h)=\frac{1}{20} t \\ & 20 \ln \left(\frac{5}{6-h}\right)=t \end{aligned}Part ii)
\begin{aligned} h=2: \quad t & =20 \ln \left(\frac{5}{6-2}\right) \\ & =4.5 \text { years. } \end{aligned}Part iii)
t=10: \quad \begin{aligned} 20 \ln \left(\frac{5}{6-h}\right) & =10 \\ \ln \frac{5}{6-h} & =\frac{1}{2} \\ e^{\frac{1}{2}} & =\frac{5}{6-h} \\ 6-h & =\frac{5}{e^{1 / 2}} \Rightarrow h=2.97 \mathrm{~m} \end{aligned}Part iv) Approximately 6m
Question 3
Solution
Part i)
\begin{aligned} \frac{d x}{d t} & =-k \sqrt{x} \\ \int \frac{1}{\sqrt{x}} d x & =-k \int d t \\ \int x^{-\frac{1}{2}} d x & =-k \int d t \\ 2 x^{\frac{1}{2}} & =-k t+c \end{aligned}Part ii)
\begin{aligned} t=0 ; & x=2: \quad 2 \sqrt{2}=c \\ & \Rightarrow 2 \sqrt{x}=-k t+2 \sqrt{2} \end{aligned}Part ii)
\begin{aligned} t=5 ; x=1 & \\ 2 & =-5 k+2 \sqrt{2} \\ k & =\frac{2 \sqrt{2}-2}{5} \\ \therefore \quad 2 \sqrt{x} & =-\left(\frac{2 \sqrt{2}-2}{5}\right) t+2 \sqrt{2} \end{aligned}\begin{gathered} x=0.5: \frac{2 \sqrt{0.5}-2 \sqrt{2}}{-\left(\frac{2 \sqrt{2}-2}{5}\right)}=t \\ \therefore t=8.5 \end{gathered}Question 4
Solution
\begin{aligned} &\frac{16+5 x-2 x^2}{(x+1)^2(x+4)}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{x+4}\\ &16+5 x-2 x^2=A(x+1)(x+4)+B(x+4)+C(x+1)^2 \end{aligned}\begin{array}{rlrl} x=-1: & 3 B & =9 \\ B & =3 \end{array}\begin{array}{rlrl} x=-4: & -36 & =9 c \\ c & =-4 \end{array}\begin{gathered} x=0: \quad 16=4 A+4 B+C \\ A=2 \end{gathered}\Rightarrow \frac{2}{x+1}+\frac{3}{(x+1)^2}-\frac{4}{x+4}Part ii)
\frac{d y}{d x}=\frac{16+5 x-2 x^2}{(x+1)^2(x+4)} y\begin{aligned} & \int \frac{1}{y} d y=\int \frac{2}{x+1}+\frac{3}{(x+1)^2}-\frac{4}{x+4} d x \\ & \ln y=2 \int \frac{1}{x+1} d x+3 \int \frac{1}{(x+1)^2} d x-4 \int \frac{1}{x+4} d x \\ & \ln y=2 \ln (x+1)-\frac{3}{x+1}-4 \ln (x+4)+c \end{aligned}\begin{aligned} & y=\frac{1}{256} \quad x=0: \\ & \ln \frac{1}{256}=2 \ln 1-\frac{3}{1}-4 \ln 4+c \\ & \ln \frac{1}{256}=-3-4 \ln 4+c \\ & \ln \frac{1}{256}+4 \ln 4+3=c \\ & \therefore c=3 \end{aligned}\Rightarrow \ln y=2 \ln (x+1)-\frac{3}{x+1}-4 \ln (x+4)+3\begin{aligned} x=2: \quad \ln y & =2 \ln 3-\frac{3}{3}-4 \ln 6+3 \\ & =2+\ln 3^2-\ln 6^4 \\ & =2+\ln 9-1296 \\ & =2+\ln \frac{9}{1296} \\ \therefore \ln y & =\ln e^2+\ln \frac{1}{144} \\ & =\ln \frac{e^2}{144} \\ \therefore y & =\frac{e^2}{144} \end{aligned}In conclusion, differential equations are an important topic studied in A Level Maths. They provide a powerful tool for modelling and understanding various real-world phenomena across different disciplines. By solving these equations, one can gain insights into the behaviour and changes of quantities over time or space.