Topics
Mathematical Logic
Matrices
Differentiation
Applications of Derivatives
Integration
Definite Integration
Applications of Definite Integration
- Standard Forms of Parabola and Their Shapes
- Ellipse and its Types
- Area Under Simple Curves
- Overview of Application of Definite Integration
Differential Equation and Applications
- Basic Concepts of Differential Equations
- Order and Degree of a Differential Equation
- Formation of Differential Equation by Eliminating Arbitary Constant
- Methods of Solving Differential Equations> Variable Separable Differential Equations
- Methods of Solving Differential Equations> Homogeneous Differential Equations
- Methods of Solving Differential Equations>Linear Differential Equations
- Applications of Differential Equation
- Overview of Differential Equations
Commission, Brokerage and Discount
- Commission and Brokerage Agent
- Concept of Discount
- Overview of Commission, Brokerage and Discount
Insurance and Annuity
- Insurance
- Types of Insurance
- Annuity
- Overview of Insurance and Annuity
Linear Regression
- Regression
- Types of Linear Regression
- Fitting Simple Linear Regression
- The Method of Least Squares
- Lines of Regression of X on Y and Y on X Or Equation of Line of Regression
- Properties of Regression Coefficients
- Overview: Linear Regression
Time Series
- Introduction to Time Series
- Uses of Time Series Analysis
- Components of a Time Series
- Mathematical Models
- Measurement of Secular Trend
- Overview of Time Series
Index Numbers
- Weighted Aggregate Method
- Cost of Living Index Number
- Method of Constructing Cost of Living Index Numbers - Aggregative Expenditure Method
- Overview of Index Numbers
- Method of Constructing Cost of Living Index Numbers - Family Budget Method
- Uses of Cost of Living Index Number
Linear Programming
Assignment Problem and Sequencing
- Assignment Problem
- Hungarian Method of Solving Assignment Problem
- Special Cases of Assignment Problem
- Sequencing Problem
- Types of Sequencing Problem
- Finding an Optimal Sequence
- Overview of Assignment Problem and Sequencing
Probability Distributions
Maharashtra State Board: Class 12
Definition: Equations in Variable Separable Form
The equation \[\frac{dy}{dx} = F(x, y)\] is said to be in variable separable form if it can be expressed as \[g(x) dx = h(y) dy\] or equivalently as \[\frac{dy}{dx} = g(x)h(y)\] so that the variables can be separated and integrated.
Method
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Write the equation in the form \[\frac{dy}{dx} = g(x)h(y)\].
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Rearrange it so that all y-terms are with dy, and all x-terms are with dx.
\[\frac{1}{h(y)} dy = g(x) dx\] -
Integrate both sides.
\[\int \frac{1}{h(y)} dy = \int g(x) dx\] -
Add the constant of integration.
\[\int \frac{1}{h(y)} dy = \int g(x) dx + C\] -
If an initial condition is given, substitute it to obtain the particular solution.
Example 1
Solve:
\[\frac{dy}{dx} = \frac{x + 1}{2 - y}\]
Solution:
Separate the variables:
Integrate both sides:
On simplification, the general solution as
Key Points: Variable Separable Differential Equations
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Variable separable equations can be rewritten as x-part = y-part.
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Separate variables first, then integrate.
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Use one constant of integration.
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Apply the initial condition only after getting the general solution.
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Final answers may be explicit or implicit.
Example Question 1
Solve the following equation.
2x + 2 = 8
2x + 2 = 8
∴ 2x + 2 - 2 = 8 - 2
∴ 2x = 6
∴ x = 3
Example Question 2
Solve the following equation.
3x - 5 = x - 17
3x - 5 = x - 17
3x - 5 + 5 - x = x - 17 + 5 - x
∴ 2x = - 12
∴ x = - 6
Example Question 3
The length of a rectangle is 1 cm more than twice its breadth. If the perimeter of the rectangle is 50 cm, find its length.
Let the breadth of the rectangle be x cm.
Then the length of the rectangle will be (2x +1)cm.
2 × length + 2 × breadth = perimeter of rectangle
2 (2x + 1) + 2x = 50
∴ 4x + 2 + 2x = 50
∴ 6x + 2 = 50
∴ 6x = 50 - 2 = 48
∴ x = 8
Breadth of rectangle is 8 cm.
Length of the rectangle = 2x + 1 = 2 × 8 + 1
∴ Length of rectangle = 17 cm.
Example Question 4
Solve the following equation.
The sum of two consecutive natural numbers is 69. Find the numbers.
Let one natural number be x.
The next natural number is x + 1
(x) + (x + 1) = 69
∴ x + x + 1 = 69
∴ 2x + 1 = 69
2x = 69 - 1
∴ 2x = 68
∴ x = 34
1st natural number = 34
2nd natural number = 34 + 1 = 35.
