Understanding the Problem: Differential Equation for Straight Lines

The question asks us to find the differential equation that describes a specific family of straight lines. This family consists of all straight lines that are exactly one unit away from the origin (the point (0,0)).

Equation of a Straight Line and Distance from Origin

A general straight line can be represented by the equation:

\(y = mx + c\)

where \(m\) is the slope and \(c\) is the y-intercept.

We can rewrite this equation in the standard form \(Ax + By + C = 0\):

\(mx - y + c = 0\)

The distance from a point \((x_0, y_0)\) to a line \(Ax + By + C = 0\) is given by the formula:

\(\text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}\)

In our case, the point is the origin \((0,0)\), and the line is \(mx - y + c = 0\). Here, \(A=m\), \(B=-1\), and \(C=c\). The distance from the origin to the line is:

\(\text{Distance} = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}} = \frac{|c|}{\sqrt{m^2 + 1}}\)

Applying the Unit Distance Condition

We are given that the distance from the origin to the line is one unit. So, we set the distance equal to 1:

\(\frac{|c|}{\sqrt{m^2 + 1}} = 1\)

This implies:

\(|c| = \sqrt{m^2 + 1}\)

Squaring both sides gives:

\(c^2 = m^2 + 1\)

And therefore, \(c = \pm \sqrt{m^2 + 1}\).

Equation of the Family of Straight Lines

Substituting the expression for \(c\) back into the equation of the straight line \(y = mx + c\), we get the equation for the family of straight lines that are at unit distance from the origin:

\(y = mx \pm \sqrt{m^2 + 1}\)

This equation contains the parameter \(m\) (the slope). To find the differential equation, we need to eliminate this parameter.

Forming the Differential Equation

We have the equation:

\(y = mx \pm \sqrt{m^2 + 1}\)

Differentiate this equation with respect to \(x\). Recall that \(m\) is a constant for a specific line but is the parameter we want to eliminate for the family of lines. \(\frac{dy}{dx}\) represents the slope of the line, which is \(m\).

\(\frac{dy}{dx} = \frac{d}{dx}(mx \pm \sqrt{m^2 + 1})\)

\(\frac{dy}{dx} = m \frac{d}{dx}(x) \pm \frac{d}{dx}(\sqrt{m^2 + 1})\)

\(\frac{dy}{dx} = m(1) \pm 0\)

\(\frac{dy}{dx} = m\)

Now, we have \(m = \frac{dy}{dx}\). Let's substitute this back into the family of lines equation \(y = mx \pm \sqrt{m^2 + 1}\):

\(y = \left(\frac{dy}{dx}\right) x \pm \sqrt{\left(\frac{dy}{dx}\right)^2 + 1}\)

To isolate the term with the square root, rearrange the equation:

\(y - x\frac{dy}{dx} = \pm \sqrt{\left(\frac{dy}{dx}\right)^2 + 1}\)

Finally, square both sides of the equation to eliminate the square root:

\(\left(y - x\frac{dy}{dx}\right)^2 = \left(\pm \sqrt{\left(\frac{dy}{dx}\right)^2 + 1}\right)^2\)

\(\left(y - x\frac{dy}{dx}\right)^2 = \left(\frac{dy}{dx}\right)^2 + 1\)

This is the differential equation representing the family of straight lines at unit distance from the origin.

Comparing with Options

Let's compare our derived differential equation with the given options:

• Option 1: \({\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\) (Incorrect)
• Option 2: \({\left( {{\rm{y}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\) (Incorrect)
• Option 3: \({\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\) (Matches our derived equation)
• Option 4: \({\left( {{\rm{y}} + {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 - {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\) (Incorrect)

The differential equation that represents the family of straight lines at unit distance from the origin is \({\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\).

Revision Table: Key Concepts for Differential Equations

Additional Information: Geometric Interpretation

The equation \(y = mx \pm \sqrt{m^2 + 1}\) can also be related to the normal form of a straight line. The normal form is \(x \cos \alpha + y \sin \alpha = p\), where \(p\) is the perpendicular distance from the origin and \(\alpha\) is the angle the normal makes with the positive x-axis.

For our family of lines, the perpendicular distance \(p\) is 1. So the equation becomes \(x \cos \alpha + y \sin \alpha = 1\).

We can relate this form to \(y = mx + c\). If \(y = mx + c\), then \(mx - y + c = 0\). Comparing this to \(x \cos \alpha + y \sin \alpha - 1 = 0\), we see that the coefficients are proportional.

\(\frac{m}{\cos \alpha} = \frac{-1}{\sin \alpha} = \frac{c}{-1}\)

From the second equality, \(c = -(-1)/\sin \alpha = 1/\sin \alpha\). From the first two, \(m = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha\).

Substitute these into \(c^2 = m^2 + 1\):

\(\left(\frac{1}{\sin \alpha}\right)^2 = (-\cot \alpha)^2 + 1\)

\(\frac{1}{\sin^2 \alpha} = \cot^2 \alpha + 1\)

\(\csc^2 \alpha = \cot^2 \alpha + 1\)

This is a trigonometric identity, confirming that the family of lines \(x \cos \alpha + y \sin \alpha = 1\) represents all lines at unit distance from the origin. The differential equation we found is the result of eliminating the parameter \(\alpha\) (or equivalently \(m = -\cot \alpha\)) from this family of lines.

The differential equation \({\left( {{\rm{y}} - {\rm{x}}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2} = 1 + {\left( {\frac{{{\rm{dy}}}}{{{\rm{dx}}}}} \right)^2}\) is a first-order non-linear differential equation.