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Question

What does the equation \(x \frac{dy}{dx}-2y= 0\) represent ?

This question was previously asked in
NDA I 2022 GAT Previous Year Paper (10-Apr-2022)
The correct answer is

A family of parabolas

The question asks us to identify the geometric representation of the differential equation \(x \frac{dy}{dx}-2y= 0\). To do this, we need to solve the given differential equation and analyze the resulting general solution.

Solving the Differential Equation

The given differential equation is a first-order differential equation:

\(x \frac{dy}{dx} - 2y = 0\)

We can rearrange this equation to separate the variables \(x\) and \(y\):

\(x \frac{dy}{dx} = 2y\)

Assuming \(x \neq 0\) and \(y \neq 0\), we can write:

\(\frac{dy}{y} = \frac{2}{x} dx\)

Now, we integrate both sides of the separated equation:

\(\int \frac{dy}{y} = \int \frac{2}{x} dx\)

Performing the integration, we get:

\(\ln|y| = 2 \ln|x| + C_1\)

where \(C_1\) is the constant of integration.

Using the properties of logarithms (\(a \ln b = \ln b^a\)), we can rewrite the right side:

\(\ln|y| = \ln|x^2| + C_1\)

Now, we can exponentiate both sides to eliminate the natural logarithm:

\(e^{\ln|y|} = e^{\ln|x^2| + C_1}\)

\(|y| = e^{\ln|x^2|} \cdot e^{C_1}\)

\(|y| = |x^2| \cdot e^{C_1}\)

Let \(C_2 = e^{C_1}\). Since \(C_1\) is an arbitrary constant, \(C_2\) is a positive constant. So,

\(|y| = C_2 |x^2|\)

Since \(x^2\) is always non-negative, \(|x^2| = x^2\). Thus,

\(|y| = C_2 x^2\)

This implies \(y = \pm C_2 x^2\). Let \(C = \pm C_2\). Since \(C_2\) is a positive constant, \(C\) can be any non-zero real number. If we also consider the case \(y=0\), which was excluded during variable separation, substituting \(y=0\) into the original differential equation gives \(x \frac{d(0)}{dx} - 2(0) = 0 - 0 = 0\), which is true for all \(x\). The solution \(y=0\) corresponds to \(y = C x^2\) when \(C=0\). Therefore, the general solution is:

\(y = C x^2\)

where \(C\) is an arbitrary constant.

Interpreting the General Solution \(y = C x^2\)

The equation \(y = C x^2\) is a familiar form in coordinate geometry.

  • If \(C \neq 0\), this equation represents a parabola with its vertex at the origin (0,0) and its axis of symmetry along the y-axis.
  • The parameter \(C\) determines the shape and direction of the parabola.
    • If \(C > 0\), the parabola opens upwards (e.g., \(y = x^2\)).
    • If \(C < 0\), the parabola opens downwards (e.g., \(y = -x^2\)).
  • If \(C = 0\), the equation becomes \(y = 0 \cdot x^2\), which simplifies to \(y = 0\). This is the equation of the x-axis, which can be considered a degenerate parabola.

Since the general solution involves an arbitrary constant \(C\), it represents a collection or family of such parabolas, all passing through the origin.

Comparing with Given Options

Let's compare our solution \(y = C x^2\) with the descriptions of the families of curves given in the options:

  • A family of straight lines: Straight lines generally have the form \(y = mx + c\) or \(Ax + By + C = 0\). Our solution \(y = C x^2\) is clearly not a straight line (unless \(C=0\), which is a special case, but the general form is parabolic).
  • A family of circles: Circles have the general equation \((x-a)^2 + (y-b)^2 = r^2\). Our solution does not match this form.
  • A family of parabolas: Parabolas can have forms like \(y = ax^2 + bx + c\), \(x = ay^2 + by + c\), or rotated versions. Our solution \(y = C x^2\) is a standard form of a parabola with vertex at the origin. This matches the description.
  • A family of ellipses: Ellipses have the general equation \(\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1\) (for axes parallel to coordinates). Our solution does not match this form.

Based on the form of the general solution \(y = C x^2\), the differential equation represents a family of parabolas.

Differential Equation General Solution Form Represents
\(x \frac{dy}{dx} - 2y = 0\) \(y = C x^2\) Family of Parabolas

Conclusion

The solution to the differential equation \(x \frac{dy}{dx} - 2y = 0\) is \(y = C x^2\), which is the equation of a parabola with vertex at the origin. The arbitrary constant \(C\) signifies that this is a family of such parabolas. Therefore, the equation represents a family of parabolas.

Revision Table: Differential Equations and Geometric Forms

Differential Equation Type Common Solutions Represents
\(dy/dx = k\) \(y = kx + C\) Family of straight lines (with same slope)
\(dy/dx = y/x\) \(y = Cx\) Family of straight lines (passing through origin)
\(dy/dx = -x/y\) \(x^2 + y^2 = C\) Family of circles (centered at origin)
\(x dy/dx - ny = 0\) \(y = Cx^n\) Family of curves (e.g., parabolas for n=2, cubic for n=3)

Additional Information: Understanding Families of Curves

A family of curves is a collection of curves related to each other by a common property. In the context of differential equations, the general solution of a first-order differential equation typically involves one arbitrary constant. As this constant varies, it generates different individual curves, collectively forming a family. For example, \(y = C x^2\) represents a family of parabolas. Each specific value of \(C\) (like \(C=1\), \(C=2\), \(C=-0.5\)) gives a unique parabola within that family.

Differential equations often arise in physical problems, and their solutions describe the behavior of systems. The general solution represents all possible states or paths (the family of curves), while a particular solution (obtained using initial conditions) represents a specific state or path within that family.

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Important Questions from Differential Equations

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    The solution of the differential equation when the source f(t) = θ(t) (the Heaviside step function) is

  2. \(\smallint \frac{{dx}}{{{{\left( {x + 1} \right)}^2}\left( {{x^2} + 1} \right)}}\) = ?
  3. 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:

  4. Complimentary function of the differential equation \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = {e^{{x^2}}}\)

  5. 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:

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