Consider the following for the next two (02) items that follow : Suppose E is the differential equation representing family of curves y2 = 2cx + 2c√c where c is a positive parameter.
What is the order of the differential equation ?
1
Understanding Differential Equations and Order
The question asks for the order of the differential equation that represents a given family of curves. The family of curves is described by the equation \(y^2 = 2cx + 2c\sqrt{c}\), where \(c\) is a positive parameter.
The order of a differential equation is defined as the order of the highest derivative appearing in the equation. To find the differential equation representing a given family of curves, we need to eliminate the arbitrary parameters present in the equation of the family.
Eliminating the Parameter to Form the Differential Equation
The given equation for the family of curves is:
\(y^2 = 2cx + 2c\sqrt{c}\)
We can rewrite \(2c\sqrt{c}\) as \(2c^{3/2}\).
\(y^2 = 2cx + 2c^{3/2}\)
This equation contains one arbitrary parameter, \(c\). To eliminate one parameter, we typically need to differentiate the equation once with respect to the independent variable \(x\).
Step-by-Step Differentiation
We differentiate the equation \(y^2 = 2cx + 2c^{3/2}\) with respect to \(x\). Remember that \(y\) is considered a function of \(x\), and \(c\) is a constant parameter.
Applying the chain rule to \(y^2\):
\(\frac{d}{dx}(y^2) = 2y \frac{dy}{dx}\)
Differentiating the term \(2cx\) with respect to \(x\):
\(\frac{d}{dx}(2cx) = 2c \frac{dx}{dx} = 2c\)
Differentiating the term \(2c^{3/2}\) with respect to \(x\):
\(\frac{d}{dx}(2c^{3/2}) = 0\), since \(c\) is a constant with respect to \(x\).
Now, we differentiate the entire equation:
\(\frac{d}{dx}(y^2) = \frac{d}{dx}(2cx) + \frac{d}{dx}(2c^{3/2})\)
\(2y \frac{dy}{dx} = 2c + 0\)
Using the notation \(y' = \frac{dy}{dx}\), this becomes:
\(2y y' = 2c\)
Dividing both sides by 2, we get an expression for \(c\) in terms of \(y\) and \(y'\):
\(c = y y'\)
Substituting Back to Obtain the Differential Equation
Now we substitute this expression for \(c\) back into the original equation of the family of curves \(y^2 = 2cx + 2c^{3/2}\) to eliminate the parameter \(c\).
Substitute \(c = yy'\):
\(y^2 = 2(yy')x + 2(yy')^{3/2}\)
\(y^2 = 2xyy' + 2(yy')\sqrt{yy'}\)
This equation \(y^2 = 2xyy' + 2(yy')^{3/2}\) is the differential equation that represents the given family of curves.
Determining the Order of the Differential Equation
The order of a differential equation is the order of the highest derivative that appears in the equation. Let's look at the equation we obtained:
\(y^2 = 2xyy' + 2(yy')^{3/2}\)
The only derivative present in this equation is \(y'\), which represents the first derivative \(\frac{dy}{dx}\).
Since the highest order derivative is the first derivative, the order of this differential equation is 1.
Conclusion: Order of the Differential Equation
By eliminating the single arbitrary parameter \(c\) from the equation of the family of curves \(y^2 = 2cx + 2c\sqrt{c}\), we obtained a differential equation where the highest derivative is the first derivative. Therefore, the order of the differential equation is 1.
Revision Table: Key Concepts in Differential Equations
| Concept | Explanation |
|---|---|
| Differential Equation | An equation that connects an independent variable, a dependent variable, and the derivatives of the dependent variable. |
| Order of Differential Equation | The rank of the highest derivative present in the differential equation. If the highest derivative is \( \frac{d^n y}{dx^n} \), the order is \(n\). |
| Family of Curves | A collection of curves whose equations depend on one or more parameters. Changing the parameter value selects a specific curve from the family. |
| Parameter Elimination | The process of removing arbitrary constants (parameters) from the equation of a family of curves, usually done by differentiation, to obtain the corresponding differential equation. |
Additional Information: Forming Differential Equations
The standard method for finding the differential equation that represents a family of curves involves differentiation. The number of times you need to differentiate is generally equal to the number of essential arbitrary constants in the family's equation.
- If a family of curves has \(n\) essential arbitrary constants, the differential equation representing it will typically be of order \(n\).
- The steps usually involve writing down the equation of the family, differentiating it repeatedly, and then using algebraic manipulation (including substituting derivatives back into the original or differentiated equations) to eliminate all the arbitrary constants.
- The resulting equation, free of arbitrary constants and involving derivatives, is the differential equation for the family.
Similar Questions
What is the differential equation of all parabolas of the type y2 = 4a (x - b)?
The solution of the differential equation \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = \frac{{{\rm{y}}\phi '\left( {\rm{x}} \right) - {{\rm{y}}^2}}}{{\phi \left( {\rm{x}} \right)}}\) is
- If y = (x x ) x , then which one of the following is correct ?
- If y = ln 2 \(\left(\frac{x^2−x+1}{x^2+x+1}\right)\) , then what is \(\frac{\text{dy}}{\text{dx}}\) at x = 0 equal to ?
What is the solution of the differential equation \(\frac{{dx}}{{dy}} = \frac{{x\; + {\rm{\;}}y\; + {\rm{\;}}1}}{{x\; + {\rm{\;}}y\; - {\rm{\;}}1}}?\)
Where c is the arbitrary constant
What is the solution of the differential equation \(\ln \left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right) = {\rm{ax}} + {\rm{by}}?\)
The equation of the curve passing through the point (-1, -2) which satisfies \(\frac{{{\rm{dy}}}}{{{\rm{dx}}}} = {\rm{}} - {{\rm{x}}^2} - \frac{1}{{{{\rm{x}}^3}}},{\rm{is}}\)
What does the equation \(x \frac{dy}{dx}-2y= 0\) represent ?
What is the solution of (1 +2x) dy – (1 – 2y) dx = 0?
Which one of the following differential equations has a periodic solution?
Important Questions from Differential Equations
The Green's function for the differential equation \(\rm\frac{d^2x}{dt^2}\) + x = f(t), satisfying the initial conditions x(0) = \(\rm\frac{dx}{dt}\) (0) = 0, is
G(t, τ) = \(\begin{cases}0 & \text { for } \quad 0<\rm t<\tau \\ \sin (\rm t−\tau) & \text { for } \quad \rm t>\tau\end{cases}\)
The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is
- \(\smallint \frac{{dx}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}}\) = ?
A differential equation is given as x (t + 2) + 3x (t + 1) + 2x (t) =0; x(0) = 0, x(1) = 1. The solution of this equation will be:
Complimentary function of the differential equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = {e^{{x^2}}}\)
If y = \(\rm\left(\frac{1}{x}\right)^x \), then value of \(\rm e^e\left(\frac{d^2 y}{d x^2}\right)_{x=e}\) is:
