Differential Equations — Class 12 Maths Notes

Topics covered in Differential Equations

1. 1.Introduction of Differential Equations

   Introduction to Differential Equations

   Imagine you're tracking the growth of a plant in your garden. Each day, you measure how much taller the plant has grown compared to the previous day. This daily growth rate, when described using mathematics, can lead us to what's known as a differential equation. Differential equations are a way to express how things change under various conditions and are used extensively in science, engineering, economics, and beyond.

   What is a Differential Equation?

   A differential equation is a mathematical equation that relates some function with its derivatives. In simple terms, it describes the rate at which something changes. When you're looking at the plant in your garden, a differential equation might model its growth by relating the rate of growth to the current height of the plant.

   Real-Life Example:

   Consider a cup of hot tea placed in your room. The rate at which the tea cools down depends on the difference in temperature between the tea and the room. A differential equation could model this cooling process, helping us predict when the tea will reach room temperature.

   Types of Differential Equations:

   1. Ordinary Differential Equations (ODEs) - These involve derivatives of a function with respect to one variable. For example, the equation describing the motion of a pendulum.
   2. Partial Differential Equations (PDEs) - These involve derivatives with respect to more than one variable. For instance, the equation modeling the temperature distribution in a room.

   Application in Careers and Industries:

   Differential equations are foundational in many fields. Engineers use them to model systems in control engineering, aerospace, and automotive industries. In biology, they model population dynamics and spread of diseases. Economists use them to model economic growth and financial markets.

   Hands-On Activity:

   Try to model the cooling of a hot beverage at room temperature. Measure the temperature of the beverage every minute for 10 minutes and see how it changes. This real-life data collection can help you understand the practical application of differential equations in everyday phenomena.

2. 2.Basic Concepts of Differential Equations

   Basic Concepts of Differential Equations

   Differential equations are fascinating mathematical tools that allow us to model changes and phenomena in various fields. Let’s break down the basic concepts:

   1. Function and Derivatives

   • Function (y): Represents the quantity that changes, like the height of a plant or the amount of money in a bank account.
   • Derivative (y'): Represents the rate of change of the function. For instance, how quickly the plant grows or how fast money accumulates interest.

   2. Order of a Differential Equation

   • The order of a differential equation is determined by the highest derivative it includes. For example, if the highest derivative is the second derivative 𝑦′′y′′, it’s a second-order differential equation.

   3. Linearity

   • A differential equation is linear if it can be expressed without multiplying the function or its derivatives together. Non-linear equations involve terms where the function and its derivatives are multiplied or have powers greater than one.

   4. Initial Conditions

   • These are values given for the function at a specific point, which are used to find a specific solution to a differential equation. For example, knowing the initial temperature of a substance helps in modeling its cooling process.

   Examples of Differential Equations

   Example 1: Population Growth

   • Equation: 𝑑𝑃𝑑𝑡=𝑘𝑃dtdP​=kP
   • Explanation: Here, 𝑃P represents the population size, and 𝑑𝑃𝑑𝑡dtdP​ is the rate of population change over time. The constant 𝑘k is the growth rate. This model assumes the population grows exponentially.

   Example 2: Radioactive Decay

   • Equation: 𝑑𝑁𝑑𝑡=−𝜆𝑁dtdN​=−λN
   • Explanation: In this model, 𝑁N represents the amount of a radioactive substance, and 𝑑𝑁𝑑𝑡dtdN​ describes how quickly the substance decays. The constant 𝜆λ is the decay constant, indicating the speed of decay, which is always negative, showing that the substance decreases over time.

   Example 3: Cooling of an Object

   • Equation: 𝑑𝑇𝑑𝑡=−𝑘(𝑇−𝑇env)dtdT​=−k(T−Tenv​)
   • Explanation: 𝑇T represents the temperature of an object, and 𝑇envTenv​ is the environmental temperature. The rate of temperature change 𝑑𝑇𝑑𝑡dtdT​ depends on the difference between the object's temperature and the environment, scaled by a constant 𝑘k. This is known as Newton's Law of Cooling.

   Applications and Real-Life Relevance

   These examples showcase how differential equations model natural phenomena and processes:

   • Population Growth: Used by ecologists and governments to predict changes in population, plan resources, and understand ecological dynamics.
   • Radioactive Decay: Important in nuclear physics, medicine (for example, in radiology), and archaeology (for carbon dating).
   • Cooling of an Object: Used in engineering to design cooling systems, in forensic science to estimate time of death, and in culinary science to ensure food safety.

   Differential equations help us understand and predict a wide range of natural and human-made phenomena, making them invaluable in science, engineering, economics, and beyond.

3. 3.Exercise Questions

   1. Equation:

   𝑑4𝑦𝑑𝑥4+sin⁡(𝑦)=0dx4d4y​+sin(y)=0

   • Type: This is a non-linear fourth-order ordinary differential equation due to the sine term involving 𝑦y.
   • Physical Context: This kind of equation could model complex mechanical systems where the displacement 𝑦y depends non-linearly on other factors, such as in non-linear springs or advanced materials under stress.

   2. Equation:

   𝑦′′+5𝑦=0y′′+5y=0

   • Type: This is a linear second-order ordinary differential equation with constant coefficients.
   • Physical Context: It typically models systems with harmonic motion, such as a mass on a spring or an LC circuit in electronics, where 𝑦y could represent displacement or electric charge.

   3. Equation:

   𝑑4𝑠𝑑𝑡4+3𝑑3𝑠𝑑𝑡3=0dt4d4s​+3dt3d3s​=0

   • Type: This is a linear fourth-order ordinary differential equation with constant coefficients.
   • Physical Context: This equation can describe systems with multiple degrees of freedom, such as mechanical vibrations in structures with damping, where 𝑠s might represent displacement.

   Solution Methods

   For Equation 1:

   Solving this equation analytically can be challenging due to the non-linear term sin⁡(𝑦)sin(y). Numerical methods or special techniques like series solutions or perturbation methods are typically used.

   For Equation 2:

   The general solution can be found using the characteristic equation:

   𝑟2+5=0 ⟹ 𝑟=±𝑖5r2+5=0⟹r=±i5​

   Thus, the solution is:

   𝑦=𝑐1cos⁡(5𝑥)+𝑐2sin⁡(5𝑥)y=c1​cos(5​x)+c2​sin(5​x)

   where 𝑐1c1​ and 𝑐2c2​ are constants determined by initial conditions.

   For Equation 3:

   The characteristic equation is:

   𝑟4+3𝑟3=0 ⟹ 𝑟(𝑟3+3𝑟2)=0 ⟹ 𝑟=0,−3r4+3r3=0⟹r(r3+3r2)=0⟹r=0,−3

   Therefore, the general solution is:

   𝑠=𝑐1+𝑐2𝑡+𝑐3𝑡2+𝑐4𝑒−3𝑡s=c1​+c2​t+c3​t2+c4​e−3t

   where 𝑐1,𝑐2,𝑐3,c1​,c2​,c3​, and 𝑐4c4​ are constants.

4. 4.General and Particular Solutions of a Differential Equation

   When solving differential equations, it's important to understand the concepts of general and particular solutions:

   General Solution:

   A general solution of a differential equation includes all possible solutions to the equation. It typically incorporates one or more arbitrary constants (C1, C2, etc.), which represent the family of solutions. The number of arbitrary constants in the general solution is equal to the order of the differential equation.

   Particular Solution:

   A particular solution is a specific instance of the general solution, obtained by setting the arbitrary constants to specific values. These values are usually determined by initial conditions or specific boundary conditions provided in the problem.

   Examples and Solutions

   Let's work through three example problems to better illustrate general and particular solutions:

   Example 1: First-Order Linear Differential Equation

   Equation: 𝑑𝑦𝑑𝑥+𝑦=𝑥dxdy​+y=x

   General Solution: First, we find the integrating factor: 𝜇(𝑥)=𝑒∫1 𝑑𝑥=𝑒𝑥μ(x)=e∫1dx=ex Multiply through by the integrating factor: 𝑒𝑥𝑑𝑦𝑑𝑥+𝑒𝑥𝑦=𝑒𝑥𝑥exdxdy​+exy=exx Integrate both sides: ∫𝑑(𝑒𝑥𝑦)=∫𝑒𝑥𝑥 𝑑𝑥∫d(exy)=∫exxdx
   𝑒𝑥𝑦=𝑒𝑥𝑥−𝑒𝑥+𝐶exy=exx−ex+C
   𝑦=𝑥−1+𝐶𝑒−𝑥y=x−1+Ce−x This is the general solution.

   Particular Solution: Suppose we are given the initial condition 𝑦(0)=2y(0)=2: 2=0−1+𝐶𝑒−02=0−1+Ce−0 𝐶=3C=3 So, the particular solution is: 𝑦=𝑥−1+3𝑒−𝑥y=x−1+3e−x

   Example 2: Second-Order Homogeneous Differential Equation

   Equation: 𝑦′′−3𝑦′+2𝑦=0y′′−3y′+2y=0

   General Solution: Find the characteristic equation: 𝑟2−3𝑟+2=0r2−3r+2=0 Factorize: (𝑟−1)(𝑟−2)=0(r−1)(r−2)=0 𝑟=1,2r=1,2 The general solution is: 𝑦=𝐶1𝑒𝑥+𝐶2𝑒2𝑥y=C1​ex+C2​e2x

   Particular Solution: Given 𝑦(0)=1y(0)=1 and 𝑦′(0)=0y′(0)=0: Using 𝑦(0)=1y(0)=1: 1=𝐶1+𝐶21=C1​+C2​ Using 𝑦′(0)=0y′(0)=0: 𝐶1𝑒0+2𝐶2𝑒0=0C1​e0+2C2​e0=0 𝐶1+2𝐶2=0C1​+2C2​=0 Solve these equations: 𝐶1=2,𝐶2=−1C1​=2,C2​=−1 So, the particular solution is: 𝑦=2𝑒𝑥−𝑒2𝑥y=2ex−e2x

   Example 3: Non-Homogeneous Linear Differential Equation

   Equation: 𝑦′′+𝑦=cos⁡(𝑥)y′′+y=cos(x)

   General Solution: The homogeneous part 𝑦ℎyh​ is solved by: 𝑦ℎ=𝐶1cos⁡(𝑥)+𝐶2sin⁡(𝑥)yh​=C1​cos(x)+C2​sin(x) A particular solution 𝑦𝑝yp​ can be guessed: 𝑦𝑝=𝐴sin⁡(𝑥)yp​=Asin(x) 𝑦𝑝′′+𝑦𝑝=−𝐴sin⁡(𝑥)+𝐴sin⁡(𝑥)=0yp′′​+yp​=−Asin(x)+Asin(x)=0 Adjust for non-homogeneous: 𝐴sin⁡(𝑥)+cos⁡(𝑥)Asin(x)+cos(x) (wrong guess, adjust to 𝐴cos⁡(𝑥)Acos(x)): 𝑦𝑝=𝐴cos⁡(𝑥)yp​=Acos(x) −𝐴cos⁡(𝑥)+𝐴cos⁡(𝑥)=cos⁡(𝑥)−Acos(x)+Acos(x)=cos(x) 𝐴=1A=1 The general solution is: 𝑦=𝐶1cos⁡(𝑥)+𝐶2𝑠𝑖𝑛(𝑥)+cos⁡(𝑥)y=C1​cos(x)+C2​sin(x)+cos(x)

   Particular Solution: With the initial condition 𝑦(0)=0y(0)=0 and 𝑦′(0)=1y′(0)=1: 0=𝐶1+10=C1​+1
   𝐶1=−1C1​=−1
   And for 𝑦′(0)=1y′(0)=1: −𝐶2+1=1−C2​+1=1
   𝐶2=0C2​=0 So, the particular solution is: 𝑦=−cos⁡(𝑥)+cos⁡(𝑥)y=−cos(x)+cos(x)

   These examples illustrate how differential equations can model dynamic systems, providing insights into both general behaviors and specific scenarios dictated by initial or boundary conditions.

5. 5.Methods of Solving First Order, First Degree Differential Equations

   Methods of Solving First Order, First Degree Differential Equations

   First order, first degree differential equations have the form: 𝑑𝑦𝑑𝑥+𝑃(𝑥)𝑦=𝑄(𝑥)dxdy​+P(x)y=Q(x) where 𝑃(𝑥)P(x) and 𝑄(𝑥)Q(x) are functions of 𝑥x. Solving these equations often involves several methods, each suitable for different types of differential equations.

   1. Separable Equations:

   These are equations where the variables can be separated into independent functions of 𝑥x and 𝑦y on different sides of the equation. They can be solved by integrating both sides.

   2. Exact Equations:

   An equation is called exact if it can be expressed as 𝑑(𝑢(𝑥,𝑦))=0d(u(x,y))=0. Solving exact equations usually involves finding a function 𝑢(𝑥,𝑦)u(x,y) such that its total differential equals the given differential equation.

   3. Integrating Factors:

   This method involves multiplying the entire differential equation by an integrating factor which is a function 𝜇(𝑥)μ(x) derived such that the resulting equation becomes exact.

   4. Homogeneous Equations:

   A first order equation is called homogeneous if it can be written in the form 𝑑𝑦𝑑𝑥=𝑓(𝑦𝑥)dxdy​=f(xy​). These are solved by substituting 𝑣=𝑦𝑥v=xy​, which transforms the differential equation into a separable form.

   5. Linear Equations:

   Linear differential equations can be directly solved by finding an integrating factor 𝜇(𝑥)μ(x), as mentioned above.

   Example Problems and Solutions

   Example 1: Separable Equation

   Equation: 𝑑𝑦𝑑𝑥=𝑦2𝑥dxdy​=xy2​

   Solution: Separate the variables: 𝑑𝑦𝑦2=𝑑𝑥𝑥y2dy​=xdx​ Integrate both sides: −1𝑦=ln⁡∣𝑥∣+𝐶−y1​=ln∣x∣+C 𝑦=−1ln⁡∣𝑥∣+𝐶y=−ln∣x∣+C1​

   Example 2: Exact Equation

   Equation: (2𝑥𝑦+𝑦2)𝑑𝑥+(𝑥2+2𝑥𝑦)𝑑𝑦=0(2xy+y2)dx+(x2+2xy)dy=0

   Solution: Check if it's exact. If not, find an integrating factor or manipulate it to become exact. Here, it is already exact, and the solution involves finding a potential function 𝑢(𝑥,𝑦)u(x,y) such that: 𝑢𝑥=2𝑥𝑦+𝑦2,𝑢𝑦=𝑥2+2𝑥𝑦ux​=2xy+y2,uy​=x2+2xy Integrate 𝑢𝑥ux​ with respect to 𝑥x and adjust for functions of 𝑦y alone: 𝑢(𝑥,𝑦)=𝑥2𝑦+𝑦2𝑥+𝑔(𝑦)u(x,y)=x2y+y2x+g(y) Differentiating with respect to 𝑦y and equating to 𝑢𝑦uy​ gives: 𝑔′(𝑦)=0g′(y)=0 So the solution is: 𝑢(𝑥,𝑦)=𝑥2𝑦+𝑦2𝑥=𝐶u(x,y)=x2y+y2x=C

   Example 3: Integrating Factor

   Equation: 𝑑𝑦𝑑𝑥+2𝑥𝑦=𝑥2dxdy​+x2​y=x2

   Solution: Find the integrating factor: 𝜇(𝑥)=𝑒∫2𝑥 𝑑𝑥=𝑥2μ(x)=e∫x2​dx=x2 Multiply through by 𝜇(𝑥)μ(x): 𝑥2𝑑𝑦𝑑𝑥+2𝑥𝑦=𝑥4x2dxdy​+2xy=x4 Integrate: ∫𝑑(𝑥2𝑦)=∫𝑥4 𝑑𝑥∫d(x2y)=∫x4dx
   𝑥2𝑦=𝑥55+𝐶x2y=5x5​+C 𝑦=𝑥35+𝐶𝑥2y=5x3​+x2C​

   Example 4: Homogeneous Equation

   Equation: 𝑑𝑦𝑑𝑥=𝑥+𝑦𝑥dxdy​=xx+y​

   Solution: Substitute 𝑣=𝑦𝑥v=xy​ so 𝑦=𝑣𝑥y=vx and 𝑑𝑦=𝑣𝑑𝑥+𝑥𝑑𝑣dy=vdx+xdv: 𝑣𝑑𝑥+𝑥𝑑𝑣𝑑𝑥=1+𝑣dxvdx+xdv​=1+v
   𝑣+𝑥𝑑𝑣𝑑𝑥=1+𝑣v+xdxdv​=1+v
   𝑥𝑑𝑣𝑑𝑥=1xdxdv​=1 Separate and integrate: 𝑑𝑣1=𝑑𝑥𝑥1dv​=xdx​ 𝑣=ln⁡∣𝑥∣+𝐶v=ln∣x∣+C 𝑦=𝑥(ln⁡∣𝑥∣+𝐶)y=x(ln∣x∣+C)

   These examples illustrate different methods for solving first order, first degree differential equations, highlighting the versatility and complexity of techniques required depending on the form of the equation

6. 6.Exercise Questions

   Problem 1:

   Given: 𝑑𝑦𝑑𝑥−3𝑦cot⁡𝑥=sin⁡2𝑥;𝑦=2 when 𝑥=𝜋2dxdy​−3ycotx=sin2x;y=2 when x=2π​

   Solution: This is a linear first-order differential equation. We can solve it using an integrating factor.

   Step 1: Find the integrating factor 𝜇(𝑥)μ(x). 𝜇(𝑥)=𝑒∫−3cot⁡𝑥 𝑑𝑥=𝑒−3log⁡∣sin⁡𝑥∣=(sin⁡𝑥)−3μ(x)=e∫−3cotxdx=e−3log∣sinx∣=(sinx)−3

   Step 2: Multiply the entire differential equation by 𝜇(𝑥)μ(x) to make it exact. (sin⁡𝑥)−3𝑑𝑦𝑑𝑥−3𝑦(sin⁡𝑥)−4cos⁡𝑥=(sin⁡𝑥)−3sin⁡2𝑥(sinx)−3dxdy​−3y(sinx)−4cosx=(sinx)−3sin2x

   Step 3: Integrate the equation. 𝑑𝑑𝑥[𝑦(sin⁡𝑥)−3]=(sin⁡𝑥)−3sin⁡2𝑥dxd​[y(sinx)−3]=(sinx)−3sin2x 𝑦(sin⁡𝑥)−3=∫(sin⁡𝑥)−3sin⁡2𝑥 𝑑𝑥+𝐶y(sinx)−3=∫(sinx)−3sin2xdx+C

   This integration on the right side may require substitution or numerical methods for a solution. For simplicity, let's note that this step involves complex integration techniques or software assistance.

   Step 4: Use the initial condition to solve for 𝐶C. Given 𝑦=2y=2 when 𝑥=𝜋2x=2π​: 2(sin⁡𝜋2)−3=𝐶2(sin2π​)−3=C 𝐶=2C=2

   Final solution: 𝑦=2(sin⁡𝑥)3+(sin⁡𝑥)3[integral part]y=2(sinx)3+(sinx)3[integral part]

   Problem 2:

   Given: Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (𝑥,𝑦)(x,y) is equal to the sum of the coordinates of the point.

   Solution: 𝑑𝑦𝑑𝑥=𝑥+𝑦dxdy​=x+y

   Step 1: Solve the differential equation using an integrating factor. 𝜇(𝑥)=𝑒∫1 𝑑𝑥=𝑒𝑥μ(x)=e∫1dx=ex 𝑒𝑥𝑑𝑦𝑑𝑥+𝑒𝑥𝑦=𝑒𝑥𝑥exdxdy​+exy=exx 𝑑𝑑𝑥(𝑒𝑥𝑦)=𝑒𝑥𝑥dxd​(exy)=exx 𝑒𝑥𝑦=∫𝑒𝑥𝑥 𝑑𝑥+𝐶exy=∫exxdx+C 𝑒𝑥𝑦=𝑒𝑥𝑥−𝑒𝑥+𝐶exy=exx−ex+C 𝑦=𝑥−1+𝐶𝑒−𝑥y=x−1+Ce−x

   Step 2: Use the initial condition (0,0)(0,0). 0=0−1+𝐶𝑒00=0−1+Ce0 𝐶=1C=1 𝑦=𝑥−1+𝑒−𝑥y=x−1+e−x

   Problem 3:

   Given: Find the equation of a curve passing through the point (0,2)(0,2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.

   Solution: 𝑑𝑦𝑑𝑥+5=𝑥+𝑦dxdy​+5=x+y
   𝑑𝑦𝑑𝑥=𝑥+𝑦−5dxdy​=x+y−5

   Step 1: Solve using an integrating factor. 𝜇(𝑥)=𝑒∫1 𝑑𝑥=𝑒𝑥μ(x)=e∫1dx=ex 𝑒𝑥𝑑𝑦𝑑𝑥+𝑒𝑥𝑦=𝑒𝑥(𝑥−5)exdxdy​+exy=ex(x−5)
   𝑑𝑑𝑥(𝑒𝑥𝑦)=𝑒𝑥𝑥−5𝑒𝑥dxd​(exy)=exx−5ex 𝑒𝑥𝑦=∫(𝑒𝑥𝑥−5𝑒𝑥) 𝑑𝑥+𝐶exy=∫(exx−5ex)dx+C
   𝑒𝑥𝑦=𝑒𝑥𝑥−5𝑒𝑥+𝐶exy=exx−5ex+C
   𝑦=𝑥−5+𝐶𝑒−𝑥y=x−5+Ce−x

   Step 2: Use the initial condition (0,2)(0,2). 2=0−5+𝐶𝑒02=0−5+Ce0 𝐶=7C=7 𝑦=𝑥−5+7𝑒−𝑥y=x−5+7e−x

   These solutions provide a detailed method for solving each differential equation based on the given initial conditions. Each problem was approached using integrating factors, a common technique for first-order linear differential equations.