The differential equation of the family of straight lines y = mx is
Differential Equation of Straight Lines: Understanding y = mx
The question asks for the differential equation of the family of straight lines given by the equation \(y = mx\).
In this equation, \(y\) and \(x\) are variables, and \(m\) is an arbitrary constant representing the slope of the line. A differential equation for a family of curves is formed by eliminating the arbitrary constants present in the equation of the family. The number of arbitrary constants determines the order of the differential equation.
Forming the Differential Equation
Let's start with the given equation of the family of straight lines:
\[y = mx \quad \ldots(1)\]
To form the differential equation, we need to eliminate the arbitrary constant \(m\). We do this by differentiating the equation with respect to \(x\).
Step 1: Differentiate the equation with respect to \(x\)
Differentiating both sides of equation (1) with respect to \(x\), we get:
\[\frac{d}{dx}(y) = \frac{d}{dx}(mx)\]
Since \(m\) is a constant, we can take it out of the differentiation:
\[\frac{dy}{dx} = m \frac{d}{dx}(x)\]
We know that \(\frac{d}{dx}(x) = 1\).
Therefore:
\[\frac{dy}{dx} = m \cdot 1\]
\[\frac{dy}{dx} = m \quad \ldots(2)\]
Analyzing the Result
Equation (2) shows that the derivative of \(y\) with respect to \(x\) is equal to \(m\), which is the constant slope of the straight line. This equation directly relates the derivative of \(y\) to the constant \(m\).
Comparing this result with the given options:
- Option 1: \(\frac{dy}{dx} = m\)
- Option 2: \(ydx - xdy = 0\)
- Option 3: \(\frac{d^2y}{dx^2} = 0\)
- Option 4: \(y dx + x dy = 0\)
The derived equation \(\frac{dy}{dx} = m\) directly matches Option 1.
Further Consideration (Eliminating m):
While \(\frac{dy}{dx} = m\) is a direct result of differentiation and directly present in the options, a complete differential equation for the family of lines typically involves eliminating the arbitrary constant \(m\) from the equations. We have:
1. \(y = mx\) (Original equation)
2. \(\frac{dy}{dx} = m\) (After differentiation)
Substituting the value of \(m\) from equation (2) into equation (1):
\[y = \left(\frac{dy}{dx}\right)x\]
This can be rearranged as:
\[y = x \frac{dy}{dx}\]
Or, in differential form:
\[y dx = x dy\]
\[x dy - y dx = 0\]
This matches Option 2 after rearranging terms, \((ydx - xdy = 0)\) or \(-(xdy - ydx) = 0\).
However, the question asks for "The differential equation of the family of straight lines \(y = mx\)" and provides \(\frac{dy}{dx} = m\) as a direct option. In such cases, the simplest form obtained by direct differentiation that matches an option is considered. The equation \(\frac{dy}{dx} = m\) clearly expresses the property of these lines (constant slope) in terms of a derivative.
Therefore, based on the given options, \(\frac{dy}{dx} = m\) is the correct choice as it represents the first derivative of the given line equation.
Summary of Steps:
| Step | Description | Equation |
|---|---|---|
| 1 | Start with the given equation of the family of lines. | \(y = mx\) |
| 2 | Differentiate with respect to \(x\). | \(\frac{dy}{dx} = m \frac{d}{dx}(x)\) |
| 3 | Simplify the derivative. | \(\frac{dy}{dx} = m\) |
| 4 | Compare with options. | Matches Option 1. |
The final answer is \(\frac{dy}{dx} = m\).
Important Questions from Formation of a Differential Equation
If x = A cos (mt - α), then the differential equation satisfying the relation is -
Which one of the following differential equations has the general solution y = ae x+ be -x ?
- Form the differential equation of $y = ae^{-2x} \cos(3x + b)$.
Where $y' = \\frac{dy}{dx}$ and $y'' = \\frac{d^2y}{dx^2}$. Which one of the following differential equation represents the family of straight lines which are at unit distance from the origin?
If y = a cos 2x + b sin 2x, then
