Samacheer Kalvi Class 12 Maths Solutions Chapter 10 Ordinary Differential Equations Exercise 10.3

Question 1. Find the differential equation of the family of (i) all non-vertical lines in a plane (ii) all non horizontal lines in a plane.
Answer:
(i) For non-vertical lines, the general equation of a family of lines in a plane is \(ax + by = 1\), where \(b \neq 0\) and \(a \in R\).
The given equation is \(ax + by = 1\) ....... (1)
We differentiate equation (1) with respect to \(x\):
\( \frac{d}{dx}(ax + by) = \frac{d}{dx}(1) \)
\( a + b \frac{dy}{dx} = 0 \)
Since there are two arbitrary constants (\(a\) and \(b\)) in the original equation, we must differentiate twice to eliminate them.
Differentiate again with respect to \(x\):
\( \frac{d}{dx}(a + b \frac{dy}{dx}) = \frac{d}{dx}(0) \)
\( 0 + b \frac{d^2y}{dx^2} = 0 \)
Since \(b \neq 0\), we can divide by \(b\):
\( \frac{d^2y}{dx^2} = 0 \)
This is the required differential equation for the family of non-vertical lines.
(ii) For non-horizontal lines, the general equation of a family of lines in a plane is \(ax + by = 1\), where \(a \neq 0\) and \(b \in R\).
The given equation is \(ax + by = 1\) ....... (1)
We differentiate equation (1) with respect to \(y\):
\( \frac{d}{dy}(ax + by) = \frac{d}{dy}(1) \)
\( a \frac{dx}{dy} + b = 0 \)
Since there are two arbitrary constants (\(a\) and \(b\)), we differentiate twice to eliminate them.
Differentiate again with respect to \(y\):
\( \frac{d}{dy}(a \frac{dx}{dy} + b) = \frac{d}{dy}(0) \)
\( a \frac{d^2x}{dy^2} = 0 \)
Since \(a \neq 0\), we can divide by \(a\):
\( \frac{d^2x}{dy^2} = 0 \)
This is the required differential equation for the family of non-horizontal lines. A second-order differential equation like this describes the general behavior of the family of curves without specific constants.
In simple words: For lines that are not straight up and down, the rate at which their slope changes (second derivative) is zero. Similarly, for lines that are not flat, their rate of change in the horizontal direction (second derivative with respect to y) is also zero.

🎯 Exam Tip: Remember to differentiate with respect to the correct variable (x for non-vertical lines, y for non-horizontal lines) and ensure all arbitrary constants are eliminated in the final differential equation.

Question 2. Form the differential equation of all straight lines touching the circle \(x^2 + y^2 = r^2\).
Answer:
The given equation of the circle is \(x^2 + y^2 = r^2\).
Let the equation of a straight line be \(y = mx + c\).
For this line to be a tangent to the circle \(x^2 + y^2 = r^2\), the condition is \(c^2 = r^2(1 + m^2)\).
Taking the square root, we get \(c = r\sqrt{1+m^2}\).
Substitute this value of \(c\) into the equation of the line:
\(y = mx + r\sqrt{1+m^2}\)
Rearrange the terms:
\(y - mx = r\sqrt{1+m^2}\) ....... (2)
We need to eliminate the arbitrary constant \(m\). Differentiate equation (2) with respect to \(x\):
\( \frac{d}{dx}(y - mx) = \frac{d}{dx}(r\sqrt{1+m^2}) \)
\( \frac{dy}{dx} - m = 0 \) (since \(r\) and \(m\) are constants when differentiating to eliminate \(m\), and \(r\sqrt{1+m^2}\) is a constant with respect to \(x\))
\( \implies m = \frac{dy}{dx} \) ....... (3)
Substitute this value of \(m\) from equation (3) back into equation (2):
\( y - x \frac{dy}{dx} = r\sqrt{1 + (\frac{dy}{dx})^2} \)
To eliminate the square root and obtain a clear differential equation, square both sides:
\( (y - x \frac{dy}{dx})^2 = r^2(1 + (\frac{dy}{dx})^2) \)
This is the required differential equation. This equation shows how the coordinates and slope of any tangent line to the given circle are related.
In simple words: We start with the math rule for a circle and a line that just touches it. We use a special property for lines that touch a circle to connect the line's slope and the circle's size. Then, we use steps to remove the slope number, which gives us a final rule that describes all such lines without needing to know their exact slope.

🎯 Exam Tip: Knowing the condition for tangency between a line and a circle is crucial. This helps in setting up the initial equation from which you can derive the differential equation by eliminating the slope parameter.

Question 3. Find the differential equation of the family of circles passing through the origin and having their centres on the x-axis.
Answer:
We are looking for circles that pass through the origin \((0,0)\) and have their centers on the x-axis. If the center is on the x-axis, its coordinates will be \((r, 0)\). Since the circle passes through the origin \((0,0)\), its radius must also be \(r\).
The general equation of a circle is \((x - h)^2 + (y - k)^2 = \text{radius}^2\).
Substituting the center \((h,k) = (r, 0)\) and radius \(r\), the equation becomes:
\((x - r)^2 + (y - 0)^2 = r^2\)
This simplifies to:
\(x^2 - 2xr + r^2 + y^2 = r^2\)
Subtracting \(r^2\) from both sides gives:
\(x^2 - 2xr + y^2 = 0\) ....... (1)
We need to eliminate the arbitrary constant \(r\). Differentiate equation (1) with respect to \(x\):
\( \frac{d}{dx}(x^2 - 2xr + y^2) = \frac{d}{dx}(0) \)
\( 2x - 2r + 2y \frac{dy}{dx} = 0 \)
Divide the entire equation by 2:
\( x - r + y \frac{dy}{dx} = 0 \)
Now, express \(r\) in terms of \(x, y,\) and \( \frac{dy}{dx} \):
\( \implies r = x + y \frac{dy}{dx} \)
Substitute this expression for \(r\) back into equation (1):
\(x^2 - 2x(x + y \frac{dy}{dx}) + y^2 = 0\)
Expand the terms:
\(x^2 - 2x^2 - 2xy \frac{dy}{dx} + y^2 = 0\)
Combine the \(x^2\) terms:
\(-x^2 - 2xy \frac{dy}{dx} + y^2 = 0\)
Multiply the entire equation by -1 to make the leading term positive:
\(x^2 + 2xy \frac{dy}{dx} - y^2 = 0\)
This is the required differential equation. This equation expresses the relationship between the coordinates of any point on such a circle and its slope.
In simple words: For circles that go through the starting point (0,0) and have their middle on the horizontal line, we first write their general math rule. Then, we use steps to remove the 'r' value (which is their size and where their center is). This gives us a final math rule that describes all such circles.

🎯 Exam Tip: When a circle passes through the origin and has its center on an axis, its radius is simply the absolute value of the coordinate of its center on that axis.

Question 4. Find the differential equation of the family of all the parabolas with latus rectum \(4a\) and whose axes are parallel to the x-axis.
Answer:
The equation for a family of parabolas with latus rectum \(4a\) and axes parallel to the x-axis, with vertex at \((a_0, b_0)\), is given by:
\((y - b_0)^2 = 4a(x - a_0)\)
Here, \(a_0\) and \(b_0\) are arbitrary constants representing the vertex coordinates. Since there are two arbitrary constants, we need to differentiate this equation twice with respect to \(x\).
Let's simplify the constants to \(a\) and \(b\), so the equation is \((y - b)^2 = 4a(x - a)\) ....... (1)
First, differentiate equation (1) with respect to \(x\):
\( \frac{d}{dx}((y - b)^2) = \frac{d}{dx}(4a(x - a)) \)
\( 2(y - b) \frac{dy}{dx} = 4a(1) \)
\( 2(y - b)y' = 4a \)
Divide by 2:
\( (y - b)y' = 2a \) ....... (2)
Next, differentiate equation (2) again with respect to \(x\). Use the product rule on the left side:
\( \frac{d}{dx}((y - b)y') = \frac{d}{dx}(2a) \)
\( y'(y') + (y - b)y'' = 0 \)
\( (y')^2 + (y - b)y'' = 0 \)
Now, we need to eliminate \(b\). From equation (2), we have \(y - b = \frac{2a}{y'}\). Substitute this into the differentiated equation:
\( (y')^2 + (\frac{2a}{y'})y'' = 0 \)
Multiply the entire equation by \(y'\) to clear the denominator:
\( (y')^3 + 2ay'' = 0 \)
This is the required differential equation. It describes the relationship between the first and second derivatives for these parabolas, showing how their slope and curvature are connected.
In simple words: For a group of parabolas whose central lines are horizontal and have a specific width, we write down their starting math rule. We then take steps to get rid of the 'a' and 'b' values, which are their constant properties. This leaves us with a final math rule that uses how the parabola bends (first and second derivatives) but not its exact position or size.

🎯 Exam Tip: The order of the differential equation must match the number of arbitrary constants in the original equation. Here, with two constants (for the vertex's x and y coordinates), a second-order differential equation is necessary.

Question 5. Find the differential equation of the family of parabolas with vertex at (0, -1) and having axis along the y axis.
Answer:
We are looking for parabolas with their vertex at \((0, -1)\) and their axis along the y-axis.
The standard equation for such a parabola, opening vertically, is \((x - h)^2 = 4a(y - k)\), where \((h,k)\) is the vertex.
Substitute the given vertex \((h,k) = (0, -1)\):
\((x - 0)^2 = 4a(y - (-1))\)
\(x^2 = 4a(y + 1)\) ....... (1)
We need to eliminate the arbitrary constant \(a\). Differentiate equation (1) with respect to \(x\):
\( \frac{d}{dx}(x^2) = \frac{d}{dx}(4a(y + 1)) \)
\( 2x = 4a \frac{d}{dx}(y + 1) \)
\( 2x = 4a y' \)
From this, we can express \(4a\):
\( 4a = \frac{2x}{y'} \)
Now, substitute this expression for \(4a\) back into equation (1):
\(x^2 = (\frac{2x}{y'})(y + 1)\)
Assuming \(x \neq 0\) (as \(x=0\) would mean \(0=4a(y+1)\) which only occurs for \(y=-1\) or \(a=0\), a degenerate case), we can divide both sides by \(x\):
\(x = \frac{2}{y'}(y + 1)\)
Multiply both sides by \(y'\):
\(xy' = 2(y + 1)\)
Expand the right side:
\(xy' = 2y + 2\)
Rearrange the terms to get the differential equation:
\(xy' - 2y - 2 = 0\)
This is the required differential equation. It describes how the slope of these parabolas relates to their x and y coordinates at any point.
In simple words: For parabolas that start at a specific point (0, -1) and open up or down along the y-axis, we write their initial math rule. Then, we take steps to get rid of the 'a' value, which controls how wide the parabola is. This gives us a final math rule that tells us how the angle of the parabola changes at any point.

🎯 Exam Tip: Always correctly identify the standard form of the conic section equation based on the given vertex and axis orientation before performing differentiation to eliminate arbitrary constants.

Question 6. Find the differential equations of the family of all the ellipses having foci on the y-axis and centre at the origin.
Answer:
For an ellipse with its center at the origin \((0,0)\) and foci on the y-axis, the major axis is along the y-axis. The equation for such an ellipse is:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) ....... (1)
Here, \(b > a\), and \(a\) and \(b\) are arbitrary constants. Since there are two arbitrary constants, we need to differentiate the equation twice with respect to \(x\).
First, differentiate equation (1) with respect to \(x\):
\( \frac{d}{dx}(\frac{x^2}{a^2} + \frac{y^2}{b^2}) = \frac{d}{dx}(1) \)
\( \frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \)
Divide the entire equation by 2:
\( \frac{x}{a^2} + \frac{y}{b^2} \frac{dy}{dx} = 0 \) ....... (2)
Next, differentiate equation (2) again with respect to \(x\). Use the product rule for the second term \( \frac{y}{b^2} \frac{dy}{dx} \):
\( \frac{d}{dx}(\frac{x}{a^2}) + \frac{d}{dx}(\frac{y}{b^2} \frac{dy}{dx}) = \frac{d}{dx}(0) \)
\( \frac{1}{a^2} + \frac{1}{b^2} [y \frac{d^2y}{dx^2} + (\frac{dy}{dx})(\frac{dy}{dx})] = 0 \)
\( \implies \frac{1}{a^2} + \frac{1}{b^2} [y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2] = 0 \) ....... (3)
From equation (2), we can express \(\frac{1}{a^2}\):
\( \frac{x}{a^2} = - \frac{y}{b^2} \frac{dy}{dx} \)
\( \implies \frac{1}{a^2} = - \frac{y}{xb^2} \frac{dy}{dx} \)
Substitute this expression for \(\frac{1}{a^2}\) into equation (3):
\( - \frac{y}{xb^2} \frac{dy}{dx} + \frac{1}{b^2} [y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2] = 0 \)
Since \(b^2 \neq 0\), we can multiply the entire equation by \(b^2\):
\( - \frac{y}{x} \frac{dy}{dx} + y \frac{d^2y}{dx^2} + (\frac{dy}{dx})^2 = 0 \)
To eliminate the denominator \(x\), multiply the entire equation by \(x\):
\( - y \frac{dy}{dx} + xy \frac{d^2y}{dx^2} + x(\frac{dy}{dx})^2 = 0 \)
Rearrange the terms to get the standard form of the differential equation:
\( xy \frac{d^2y}{dx^2} + x(\frac{dy}{dx})^2 - y \frac{dy}{dx} = 0 \)
This is the required differential equation. It describes how the curvature and slope of such ellipses are related to their position.
In simple words: For a group of ellipses centered at (0,0) with their long axis along the y-axis, we start with their standard formula. Since there are two unknown constants (a and b), we find how the ellipse changes twice. Then, we combine these equations to remove the constants, giving us a final math rule that tells us how the change rates are connected for all these ellipses.

🎯 Exam Tip: When eliminating multiple arbitrary constants, you will typically need to differentiate the original equation as many times as there are constants. Then, use algebraic substitution or elimination to remove the constants from the resulting equations.

Question 7. Find the differential equation corresponding to the family of curves represented by the equation \(y = Ae^{8x} + Be^{-8x}\), where A and B are arbitrary constants.
Answer:
The given equation for the family of curves is:
\(y = Ae^{8x} + Be^{-8x}\) ....... (1)
Since there are two arbitrary constants, \(A\) and \(B\), we need to differentiate this equation twice with respect to \(x\).
First, differentiate equation (1) with respect to \(x\):
\( \frac{dy}{dx} = \frac{d}{dx}(Ae^{8x} + Be^{-8x}) \)
\( \frac{dy}{dx} = A(8e^{8x}) + B(-8e^{-8x}) \)
\( \frac{dy}{dx} = 8Ae^{8x} - 8Be^{-8x} \) ....... (2)
Next, differentiate equation (2) again with respect to \(x\) to find the second derivative:
\( \frac{d^2y}{dx^2} = \frac{d}{dx}(8Ae^{8x} - 8Be^{-8x}) \)
\( \frac{d^2y}{dx^2} = 8A(8e^{8x}) - 8B(-8e^{-8x}) \)
\( \frac{d^2y}{dx^2} = 64Ae^{8x} + 64Be^{-8x} \)
Factor out 64 from the right side:
\( \frac{d^2y}{dx^2} = 64(Ae^{8x} + Be^{-8x}) \) ....... (3)
Now, observe that the expression in the parenthesis, \( (Ae^{8x} + Be^{-8x}) \), is exactly equal to \(y\) from the original equation (1).
Substitute \(y\) from equation (1) into equation (3):
\( \frac{d^2y}{dx^2} = 64y \)
Rearrange the terms to get the differential equation in a standard form:
\( \frac{d^2y}{dx^2} - 64y = 0 \)
This is the required differential equation. It describes a direct relationship between the curve's value and its second derivative, making it a powerful tool in many scientific fields.
In simple words: We start with a formula for curves that has two unknown numbers (A and B). To get a rule that works for all of them, we find how the curve changes, and then how that change itself changes. By looking closely, we can replace the original curve's formula back into the second change rule. This gives us a simple rule that connects the curve's height to its second rate of change.

🎯 Exam Tip: When the original equation or parts of it reappear during differentiation, always look for opportunities to substitute them back. This simplifies the process of eliminating arbitrary constants.

Question 8. Find the differential equation of the curve represented by \(xy = ae^x + be^{-x} + x^2\).
Answer:
The given equation for the curve is:
\(xy = ae^x + be^{-x} + x^2\) ....... (1)
Here, \(a\) and \(b\) are arbitrary constants. Since there are two arbitrary constants, we need to differentiate this equation twice with respect to \(x\).
First, differentiate equation (1) with respect to \(x\). Use the product rule for \(xy\):
\( \frac{d}{dx}(xy) = \frac{d}{dx}(ae^x + be^{-x} + x^2) \)
\(x \frac{dy}{dx} + y(1) = ae^x - be^{-x} + 2x\)
\( x y' + y = ae^x - be^{-x} + 2x \) ....... (2)
Next, differentiate equation (2) again with respect to \(x\). Use the product rule for \(xy'\):
\( \frac{d}{dx}(xy' + y) = \frac{d}{dx}(ae^x - be^{-x} + 2x) \)
\( (x \frac{d}{dx}(y') + y'(1)) + y' = ae^x + be^{-x} + 2 \)
\( x y'' + y' + y' = ae^x + be^{-x} + 2 \)
\( x y'' + 2y' = ae^x + be^{-x} + 2 \) ....... (3)
Now, we need to eliminate \(ae^x + be^{-x}\). From the original equation (1), we can isolate this term:
\(ae^x + be^{-x} = xy - x^2\) ....... (4)
Substitute this expression for \(ae^x + be^{-x}\) from equation (4) into equation (3):
\(x y'' + 2y' = (xy - x^2) + 2\)
Rearrange the terms to form the differential equation:
\(x y'' + 2y' - xy + x^2 - 2 = 0\)
This is the required differential equation. This equation shows the relationship between the curve's coordinates and its first and second derivatives, providing a general rule for all such curves.
In simple words: We begin with a math formula for a curve that has two unknown numbers (a and b). To find a general rule for all such curves, we apply the process of finding how the curve changes, and then how that change itself changes, twice. We also use the original formula to get rid of the unknown 'a' and 'b' terms. This leaves us with a final math rule that connects the curve's position to its bending and slope.

🎯 Exam Tip: Be meticulous with the product rule when differentiating terms involving products of variables and their derivatives. Always look for ways to substitute back the original equation to eliminate arbitrary constants effectively.