Topics
Beta and Gamma Functions, Differentiation Under Integral Sign and Exact Differential Equation old
Beta and Gamma Functions and Its Properties
Rectification of Plane Curves
Differential Equation of First Order and First Degree
Differential Calculus old
Linear Differential Eqaution with Constant Coeffiecient
Linear Differential Equations(Review), Equation Reduciable to Linear Form, Bernoulli’S Equation
Cauchy’S Homogeneous Linear Differential Equation and Legendre’S Differential Equation, Method of Variation of Parameters
Simple Application of Differential Equation of First Order and Second Order to Electrical and Mechanical Engineering Problem
Numerical Solution of Ordinary Differential Equations of First Order and First Degree and Multiple Integrals old
Multiple Integrals‐Double Integration
Taylor’S Series Method,Euler’S Method,Modified Euler Method,Runga‐Kutta Fourth Order Formula
Multiple Integrals with Application and Numerical Integration old
Triple Integration
Application to Double Integrals to Compute Area, Mass, Volume. Application of Triple Integral to Compute Volume
Numerical Integration
Differential Equations of First Order and First Degree
- Exact Differential Equations
- Equations Reducible to Exact Form by Using Integrating Factors
- Equations Reducible to Linear Equations
- Simple Application of Differential Equation of First Order and First Degree to Electrical and Mechanical Engineering Problem
Linear Differential Equations with Constant Coefficients and Variable Coefficients of Higher Order
- Linear Differential Equation with Constant Coefficient‐ Complementary Function
- Particular Integrals of Differential Equation
- Cauchy’S Homogeneous Linear Differential Equation
- Legendre’S Differential Equation
- Method of Variation of Parameters
Numerical Solution of Ordinary Differential Equations of First Order and First Degree, Beta and Gamma Function
- Taylor’S Series Method
- Euler’S Method
- Modified Euler Method
- Runga‐Kutta Fourth Order Formula
- Beta and Gamma Functions and Its Properties
Differentiation Under Integral Sign, Numerical Integration and Rectification
- Differentiation Under Integral Sign with Constant Limits of Integration
- Numerical Integration‐ by Trapezoidal
- Numerical Integration‐ by Simpson’S 1/3rd
- Numerical Integration‐ by Simpson’S 3/8th Rule
- Rectification of Plane Curves
Double Integration
- Double Integration‐Definition
- Evaluation of Double Integrals
- Change the Order of Integration
- Evaluation of Double Integrals by Changing the Order of Integration and Changing to Polar Form
Triple Integration and Applications of Multiple Integrals
- Triple Integration Definition and Evaluation
- Application of Double Integrals to Compute Area
- Application of Double Integrals to Compute Mass
- Application of Double Integrals to Compute Volume
- Application of Triple Integral to Compute Volume
Introduction
Some equations contain terms such as:
\[ \frac{1}{x}, \quad \frac{1}{y}, \quad \frac{1}{x + y}, \quad \frac{1}{x - y} \]
These are not linear in \[ x \] and \[ y. \] However, by replacing each repeated expression with a new variable, they can often be converted into simple linear equations.
Definition: Equations Reducible to Linear Equations
A pair of equations that is not linear in the original variables xxx and yyy, but becomes a pair of linear equations after a suitable substitution, is called a pair of equations reducible to linear equations.
Example
\[ \frac{4}{x} + \frac{5}{y} = 7; \quad \frac{3}{x} + \frac{4}{y} = 5 \]
Step 1: Rewrite the equations with separate coefficients
\[ 4 \left( \frac{1}{x} \right) + 5 \left( \frac{1}{y} \right) = 7 \ldots (I) \]
\[ 3 \left( \frac{1}{x} \right) + 4 \left( \frac{1}{y} \right) = 5 \ldots (II) \]
Step 2: Substitute new variables
Put \[ \frac{1}{x} = m \] and \[ \frac{1}{y} = n \]
\[ 4m + 5n = 7 \ldots (III) \]
\[ 3m + 4n = 5 \ldots (IV) \]
Step 3: Solve equations (III) and (IV)
Solving these simultaneously gives: \[ m = 3 \] and \[ n = -1 \]
Step 4: Re-substitute to find original variables \[ x \] and \[ y \]
Since \[ m = \frac{1}{x}, \] we get \[ 3 = \frac{1}{x} \implies x = \frac{1}{3} \]
Since \[ n = \frac{1}{y}, \] we get \[ -1 = \frac{1}{y} \implies y = -1 \]
Final Answer:
Solution is \[ (x, y) = \left( \frac{1}{3}, -1 \right) \]
Activity: Complete the Table
\[ \def\arraystretch{2} \begin{array}{|c|c|c|} \hline \text{Equation} & \text{No. of variables} & \text{whether linear or not} \\ \hline \dfrac{3}{x} - \dfrac{4}{y} = 8 & 2 & \text{Not linear} \\ \hline \dfrac{6}{x - 1} + \dfrac{3}{y - 2} = 0 & 2 & \text{Not linear} \\ \hline \dfrac{7}{2x + 1} + \dfrac{13}{y + 2} = 0 & 2 & \text{Not linear} \\ \hline \dfrac{14}{x + y} + \dfrac{3}{x - y} = 5 & 2 & \text{Not linear} \\ \hline \end{array} \]
Key Points: Equations Reducible to Linear Equations
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Some equations are not linear in the given variables.
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By a suitable change of variables, they can be reduced to linear equations.
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After substitution, the equations become linear in the new variables.
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Denominators must not be zero.
Example Question 1
Solve: `(x + 1)/(2x + 3) = 3/8`.
`(x + 1)/(2x + 3) = 3/8`.
We multiply both sides of the equation by (2x + 3),
`((x + 1)/(2x + 3)) xx (2x + 3) = 3/8 xx (2x + 3)`
(2x + 3) gets cancelled on the LHS we have then,
`x + 1 = (3(2x + 3))/8`
Example Question 2
Let the present ages of Anu and Raj be 4x years and 5x years respectively.
Therefore, the ratio of their ages after eight years = `(4x + 8)/(5x + 8)`
This is given to be 5: 6
Therefore,
`(4x + 8)/(5x + 8) = 5/6`
Cross-multiplication gives
6(4x + 8) = 5(5x + 8)
or 24x + 48 = 25x + 40
or 24x + 48 - 40 = 25x
or 24x + 8 = 25x
or 8 = 25x - 24x
or 8 = x
Therefore,
Anu's present age = 4x = 4 × 8 = 32 years.
Raj's present age = 5x = 5 × 8 = 40 years.
