RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.2

Find the Differential Equation of the Following Groups:

Question 1. Find the differential equation for \( y = ax + \frac{b}{x} \)
Answer: Given the equation \( y = ax + \frac{b}{x} \). Let this be equation (1). We need to remove the arbitrary constants 'a' and 'b'. First, differentiate equation (1) with respect to \( x \): \( \frac{dy}{dx} = a - \frac{b}{x^2} \) From the original equation (1), we know that \( \frac{b}{x} = y - ax \). So, \( \frac{b}{x^2} = \frac{1}{x} (y - ax) \). Now, substitute this into the first derivative: \( \frac{dy}{dx} = a - \frac{1}{x}(y - ax) \) \( \frac{dy}{dx} = a - \frac{y}{x} + a \) \( \frac{dy}{dx} = 2a - \frac{y}{x} \) Multiply the entire equation by \( x \) to clear the fraction: \( x\frac{dy}{dx} = 2ax - y \) Rearrange the terms to solve for \( 2a \): \( 2a = y + x\frac{dy}{dx} \) Now, differentiate the expression \( \frac{dy}{dx} = 2a - \frac{y}{x} \) with respect to \( x \) again: \( \frac{d^2y}{dx^2} = \frac{d}{dx}(2a) - \frac{d}{dx}\left(\frac{y}{x}\right) \) Since \( 2a \) is a constant, its derivative is 0. For \( \frac{y}{x} \), we use the quotient rule: \( \frac{d^2y}{dx^2} = 0 - \left(\frac{x\frac{dy}{dx} - y \cdot 1}{x^2}\right) \) \( \implies \frac{d^2y}{dx^2} = - \frac{x\frac{dy}{dx} - y}{x^2} \) Multiply both sides by \( x^2 \): \( x^2\frac{d^2y}{dx^2} = - (x\frac{dy}{dx} - y) \) \( \implies x^2\frac{d^2y}{dx^2} = y - x\frac{dy}{dx} \) Rearranging the terms, we get the final differential equation: \( x^2\frac{d^2y}{dx^2} + x\frac{dy}{dx} - y = 0 \) This equation no longer contains 'a' or 'b', successfully eliminating the arbitrary constants. It shows the relationship between 'y', 'x', and their derivatives.
In simple words: We start with the given equation and take its derivative two times. Each time we differentiate, we cleverly use the original equation to get rid of the constant values 'a' and 'b'. After two steps of differentiation and some rearranging, we get a new equation that only has 'x', 'y', and their derivatives.

🎯 Exam Tip: When forming a differential equation, the goal is to eliminate all arbitrary constants by differentiating the given relation as many times as there are constants. If there are two constants, differentiate twice.

Question 2. Find the differential equation for \( x^2 + y^2 = a^2 \)
Answer: Given the equation \( x^2 + y^2 = a^2 \). Here, 'a' is an arbitrary constant. We need to eliminate this constant. Differentiate the given equation with respect to \( x \): \( \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(a^2) \) Using the power rule for \( x^2 \) and the chain rule for \( y^2 \) (since \( y \) is a function of \( x \)): \( 2x + 2y\frac{dy}{dx} = 0 \) Since \( a^2 \) is a constant, its derivative is 0. Now, divide the entire equation by 2 to simplify it: \( x + y\frac{dy}{dx} = 0 \) This is the required differential equation, as the arbitrary constant 'a' has been successfully eliminated. This equation represents a family of circles centered at the origin.
In simple words: We have an equation with an 'a' in it. We take the derivative of the whole equation with respect to 'x' to get rid of 'a', because 'a' is a constant. After that, we simplify the result to find our differential equation.

🎯 Exam Tip: Remember to differentiate constants like \( a^2 \) to zero when they are standalone terms. Use the chain rule for terms involving 'y' like \( y^2 \), which becomes \( 2y\frac{dy}{dx} \).

Question 3. Find the differential equation for \( y = Ae^{3x} + Be^{5x} \)
Answer: Given the equation \( y = Ae^{3x} + Be^{5x} \). Let this be equation (1). We have two arbitrary constants, 'A' and 'B', so we will need to differentiate the equation twice to eliminate them. First, differentiate equation (1) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx}(Ae^{3x}) + \frac{d}{dx}(Be^{5x}) \) \( \frac{dy}{dx} = 3Ae^{3x} + 5Be^{5x} \) (Equation 2) Next, differentiate equation (2) with respect to \( x \) again: \( \frac{d^2y}{dx^2} = \frac{d}{dx}(3Ae^{3x}) + \frac{d}{dx}(5Be^{5x}) \) \( \frac{d^2y}{dx^2} = 9Ae^{3x} + 25Be^{5x} \) (Equation 3) Now we need to eliminate 'A' and 'B' from equations (1), (2), and (3). Multiply equation (1) by 3: \( 3y = 3Ae^{3x} + 3Be^{5x} \) (Equation 4) Subtract equation (4) from equation (2): \( \frac{dy}{dx} - 3y = (3Ae^{3x} + 5Be^{5x}) - (3Ae^{3x} + 3Be^{5x}) \) \( \frac{dy}{dx} - 3y = 2Be^{5x} \) (Equation 5) Now, differentiate equation (5) with respect to \( x \): \( \frac{d}{dx}\left(\frac{dy}{dx} - 3y\right) = \frac{d}{dx}(2Be^{5x}) \) \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} = 10Be^{5x} \) (Equation 6) From equation (5), we can express \( Be^{5x} \) as \( Be^{5x} = \frac{1}{2}\left(\frac{dy}{dx} - 3y\right) \). Substitute this expression for \( Be^{5x} \) into equation (6): \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} = 10 \cdot \frac{1}{2}\left(\frac{dy}{dx} - 3y\right) \) \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} = 5\left(\frac{dy}{dx} - 3y\right) \) \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} = 5\frac{dy}{dx} - 15y \) Bring all terms to one side to form the differential equation: \( \frac{d^2y}{dx^2} - 3\frac{dy}{dx} - 5\frac{dy}{dx} + 15y = 0 \) \( \frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0 \) This is the required second-order linear homogeneous differential equation. Such equations are fundamental in many areas of physics and engineering, describing systems like oscillations.
In simple words: We have an equation with two unknown constants. We take the derivative of the equation two times. Then, we use these three equations (the original and its two derivatives) to combine them in a way that gets rid of the two constants 'A' and 'B'. This leaves us with a differential equation that describes how 'y' changes with 'x'.

🎯 Exam Tip: For equations with exponential terms and constants, differentiate once for each constant. Then, use algebraic methods (like substitution or elimination) to remove the constants from the original equation and its derivatives. This is a common technique for solving such problems.

Question 4. Find the differential equation for \( y = e^x [A \cos x + B \sin x] \)
Answer: Given the equation \( y = e^x [A \cos x + B \sin x] \). Let this be equation (1). We need to eliminate the two arbitrary constants 'A' and 'B'. First, differentiate equation (1) with respect to \( x \). We use the product rule \( \frac{d}{dx}(uv) = u'v + uv' \): Here, \( u = e^x \) and \( v = A \cos x + B \sin x \). \( \frac{dy}{dx} = (e^x)(A \cos x + B \sin x) + e^x(-A \sin x + B \cos x) \) Using equation (1), we can substitute \( y \) for the term \( e^x(A \cos x + B \sin x) \): \( \frac{dy}{dx} = y + e^x(-A \sin x + B \cos x) \) Rearrange this to isolate the term containing the constants: \( \frac{dy}{dx} - y = e^x(-A \sin x + B \cos x) \) (Let this be equation (2)) Now, differentiate equation (2) with respect to \( x \) again, using the product rule on the right side: \( \frac{d}{dx}\left(\frac{dy}{dx} - y\right) = \frac{d}{dx}\left(e^x(-A \sin x + B \cos x)\right) \) \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = (e^x)(-A \sin x + B \cos x) + e^x(-A \cos x - B \sin x) \) Now, substitute \( \left(\frac{dy}{dx} - y\right) \) for \( e^x(-A \sin x + B \cos x) \) from equation (2): \( \frac{d^2y}{dx^2} - \frac{dy}{dx} = \left(\frac{dy}{dx} - y\right) + e^x(-A \cos x - B \sin x) \) Rearrange the terms to get the expression with constants on one side: \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^x(-A \cos x - B \sin x) \) (Let this be equation (3)) From equation (1), we know that \( A \cos x + B \sin x = \frac{y}{e^x} \). Therefore, \( -A \cos x - B \sin x = -\frac{y}{e^x} \). Substitute this into equation (3): \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = e^x\left(-\frac{y}{e^x}\right) \) \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = -y \) Bring the \( -y \) term to the left side: \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y + y = 0 \) \( \frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0 \) This is the required differential equation. It describes how the function \( y \) changes based on its first and second derivatives.
In simple words: We start with an equation that has two unknown constants 'A' and 'B'. We take the derivative two times, using the product rule carefully. After each step, we substitute back parts of the original equation or the previous derivative to get rid of 'A' and 'B'. The final equation will only contain 'y' and its derivatives.

🎯 Exam Tip: When terms like \( e^x \) are involved, look for opportunities to substitute back the original function or a previous derivative to simplify the expressions and eliminate constants more quickly. This reduces the algebraic complexity.

Question 5. Find the differential equation for \( y = a \cos (x + b) \), where 'a' and 'b' are arbitrary constants.
Answer: Given the equation \( y = a \cos (x + b) \). Let this be equation (1). We have two arbitrary constants, 'a' and 'b', so we need to differentiate the equation twice to eliminate them. First, differentiate equation (1) with respect to \( x \): \( \frac{dy}{dx} = \frac{d}{dx}(a \cos (x + b)) \) Using the chain rule, the derivative of \( \cos(u) \) is \( -\sin(u) \frac{du}{dx} \): \( \frac{dy}{dx} = -a \sin (x + b) \cdot \frac{d}{dx}(x + b) \) \( \frac{dy}{dx} = -a \sin (x + b) \) (Equation 2) Next, differentiate equation (2) with respect to \( x \) again: \( \frac{d^2y}{dx^2} = \frac{d}{dx}(-a \sin (x + b)) \) Using the chain rule again, the derivative of \( \sin(u) \) is \( \cos(u) \frac{du}{dx} \): \( \frac{d^2y}{dx^2} = -a \cos (x + b) \cdot \frac{d}{dx}(x + b) \) \( \frac{d^2y}{dx^2} = -a \cos (x + b) \) (Equation 3) Now, look back at equation (1): \( y = a \cos (x + b) \). We can directly substitute \( y \) into equation (3): \( \frac{d^2y}{dx^2} = -y \) Bring the \( -y \) term to the left side to form the differential equation: \( \frac{d^2y}{dx^2} + y = 0 \) This is the required differential equation, free from the constants 'a' and 'b'. This form often describes simple harmonic motion, like a mass on a spring or a pendulum swing.
In simple words: We have an equation with two unknown constants 'a' and 'b'. We take the derivative of the equation two times. After the second derivative, we notice that the original equation for 'y' appears again, allowing us to substitute it back. This removes the constants and gives us the final differential equation.

🎯 Exam Tip: For trigonometric functions with constants, differentiate step-by-step. Often, the original function or its first derivative will reappear in a later derivative, allowing for a straightforward substitution to eliminate the constants and simplify the process.