The differential equation of the family of curves y = p cos (ax) + q sin (ax), where p, q are arbitrary constants, is
Finding the Differential Equation for a Family of Curves
The question asks us to find the differential equation that represents the given family of curves. A family of curves is a set of curves defined by an equation containing one or more arbitrary constants. The differential equation of a family of curves is obtained by eliminating these arbitrary constants through differentiation.
The given family of curves is:
\[ y = p \cos (ax) + q \sin (ax) \]Here, \(p\) and \(q\) are arbitrary constants, while \(a\) is a fixed constant. Since there are two arbitrary constants (\(p\) and \(q\)), the differential equation will be of the second order.
We need to differentiate the given equation twice with respect to \(x\) to obtain enough equations to eliminate \(p\) and \(q\).
Step-by-Step Derivation
Step 1: Start with the original equation.
\[ y = p \cos (ax) + q \sin (ax) \quad \ldots (1) \]Step 2: Differentiate equation (1) with respect to \(x\).
Using the chain rule, the derivative of \(\cos(ax)\) is \(-a \sin(ax)\) and the derivative of \(\sin(ax)\) is \(a \cos(ax)\).
\[ \frac{dy}{dx} = \frac{d}{dx}(p \cos (ax) + q \sin (ax)) \] \[ \frac{dy}{dx} = p(-a \sin (ax)) + q(a \cos (ax)) \] \[ \frac{dy}{dx} = -ap \sin (ax) + aq \cos (ax) \quad \ldots (2) \]Step 3: Differentiate equation (2) with respect to \(x\) to find the second derivative.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(-ap \sin (ax) + aq \cos (ax)) \] \[ \frac{d^2y}{dx^2} = -ap(a \cos (ax)) + aq(-a \sin (ax)) \] \[ \frac{d^2y}{dx^2} = -a^2 p \cos (ax) - a^2 q \sin (ax) \]Step 4: Factor out \(-a^2\) from the second derivative.
\[ \frac{d^2y}{dx^2} = -a^2 (p \cos (ax) + q \sin (ax)) \]Step 5: Substitute the original expression for \(y\) from equation (1) into the equation from Step 4.
From equation (1), we know that \(y = p \cos (ax) + q \sin (ax)\). Substituting this into the equation for the second derivative:
\[ \frac{d^2y}{dx^2} = -a^2 y \]Step 6: Rearrange the equation to obtain the differential equation.
Move the term \(-a^2 y\) to the left side of the equation:
\[ \frac{d^2y}{dx^2} + a^2 y = 0 \]This is the required differential equation for the given family of curves.
Analyzing the Options
Let's compare our derived differential equation with the given options:
- Option 1: \( \frac{d^2y}{dx^2} - a^2y = 0 \)
- Option 2: \( \frac{d^2y}{dx^2} - ay = 0 \)
- Option 3: \( \frac{d^2y}{dx^2} + ay = 0 \)
- Option 4: \( \frac{d^2y}{dx^2} + a^2y = 0 \)
Our derived equation \( \frac{d^2y}{dx^2} + a^2 y = 0 \) exactly matches Option 4.
Understanding Families of Curves and Differential Equations
A differential equation represents a relationship between a function and its derivatives. When we talk about a family of curves defined by an equation with arbitrary constants, the differential equation represents a property that all curves in that family share, independent of the specific values of the constants.
The order of the differential equation needed to represent a family of curves is equal to the number of essential arbitrary constants in the equation of the family.
| Concept | Description |
|---|---|
| Family of Curves | A set of curves defined by an equation containing one or more arbitrary constants (parameters). |
| Arbitrary Constants | Parameters in the equation of a family of curves that can take any value, leading to different individual curves within the family. |
| Differential Equation | An equation relating an unknown function to its derivatives. It describes how a quantity changes. |
| Order of Differential Equation | The order of the highest derivative present in the differential equation. It equals the number of arbitrary constants eliminated. |
Revision Table: Differential Equation Formation
| Step | Action | Purpose |
|---|---|---|
| 1 | Start with the equation of the family of curves. | Establish the initial relationship. |
| 2 | Identify the number of arbitrary constants. | Determine the order of the differential equation. |
| 3 | Differentiate the equation that many times. | Create a system of equations involving derivatives and constants. |
| 4 | Eliminate the arbitrary constants from the system of equations. | Obtain an equation involving only the function and its derivatives. |
Additional Information: Homogeneous Linear Second-Order Differential Equations
The differential equation we derived, \( \frac{d^2y}{dx^2} + a^2 y = 0 \), is a specific type of second-order linear homogeneous differential equation with constant coefficients. Equations of the form \( A \frac{d^2y}{dx^2} + B \frac{dy}{dx} + Cy = 0 \) where \(A, B, C\) are constants are solved using a characteristic equation.
For \( \frac{d^2y}{dx^2} + a^2 y = 0 \), the characteristic equation is \( m^2 + a^2 = 0 \). The roots are \( m^2 = -a^2 \), which gives \( m = \pm \sqrt{-a^2} = \pm ai \).
Since the roots are complex conjugates of the form \( \alpha \pm \beta i \) (here \( \alpha = 0 \) and \( \beta = a \)), the general solution is given by:
\[ y(x) = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x)) \]Substituting \( \alpha = 0 \) and \( \beta = a \):
\[ y(x) = e^{0x} (C_1 \cos(ax) + C_2 \sin(ax)) \] \[ y(x) = 1 (C_1 \cos(ax) + C_2 \sin(ax)) \] \[ y(x) = C_1 \cos(ax) + C_2 \sin(ax) \]This matches the original family of curves \( y = p \cos (ax) + q \sin (ax) \), where \(C_1\) corresponds to \(p\) and \(C_2\) corresponds to \(q\). This confirms that the differential equation we found is indeed the one that generates this family of curves.
Similar Questions
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The degree of the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} - {\rm{x}} = {\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^{ - 4}}\) is
What is the degree of the differential equation? \(1+\left(\frac{dy}{dx}\right)^2 =\left(\frac{d^2y}{dx^2}\right)^{\frac{4}{3}}?\)
What is the order of the differential equation of all ellipses whose axes are along the coordinate axes?
What is the degree of the differential equation of all circles touching both the coordinate axes in the first quadrant?
What is the degree of the differential equation \(\frac{{{d}^{3}}y}{d{{x}^{3}}}+{{\left( \frac{dy}{dx} \right)}^{2}}-{{x}^{2}}\left( \frac{{{d}^{4}}y}{d{{x}^{4}}} \right)=0?\)
The order and degree of the differential equation y 2= 4a (x – a), where ‘a’ is an arbitrary constant, are respectively
Consider the following statements :
1. The degree of the differential equation \(\frac{\text{dy}}{\text{dx}} + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 1.
2. The order of the differential equation \(\left(\frac{\text{d}^2\text{y}}{\text{dx}^2}\right)^3 + \cos \left(\frac{\text{dy}}{\text{dx}}\right)\) = 0 is 2.
Which of the statements given above is/are correct?
What are the order and degree, respectively, of the differential equation \({\left( {\frac{{{{\rm{d}}^3}{\rm{y}}}}{{{\rm{d}}{{\rm{x}}^3}}}} \right)^2} = {{\rm{y}}^4} + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^5}?\)
What are the order and degree respectively of the differential equation whose solution is y = cx + c 2– 3c 3/2 + 2, where c is a parameter?
Important Questions from Order and Degree of a Differential Equation
Consider the following in respect of the differential equation:
\(\frac{{{d^2}y}}{{d{x^2}}} + 2{\left( {\frac{{dy}}{{dx}}} \right)^2} + 9y = x\)
1. The degree of the differential equation is 1.
2. The order of the differential equation is 2.
Which of the above statements is/are correct?
The partial differential equation \(\frac{{\partial u}}{{\partial t}} + u\frac{{\partial u}}{{\partial x}} = \frac{{{\partial ^2}u}}{{\partial {x^2}}}\) is a
The degree of the differential equation \({\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\frac{{dy}}{{dx}}} \right)^2} + \sin x\left( {\frac{{dy}}{{dx}}} \right) + y = 0\) is:
In the following partial differential equation, θ is a function of t and z, and D and K are functions of θ
\(D\left( \theta \right)\frac{{{\delta ^2}\theta }}{{\delta {z^2}}} + \frac{{\delta K\left( \theta \right)}}{{\delta z}} - \frac{{\delta \theta }}{{\delta t}} = 0\)
The above equation isThe solution of the equation \({\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}} + {\rm{y}} = 0{\rm{}}\) passing through the point (1,1) is
