Problem 2 - A Complete Resource Book in Mathematics for JEE Main

Question

The family passing through \((0,0)\) and satisfying the differential equation \(\frac{y_{2}}{y_{1}}=1\) (where \(\left.y_{n}=\frac{d^{n} y}{d x^{n}}\right)\) is (A) \(y=k\) (B) \(y=k x\) (C) \(y=k\left(e^{x}+1\right)\) (D) \(y=k\left(e^{x}-1\right)\)

Step-by-Step Solution

Answer

The solution is (D) \( y = k(e^x - 1) \).

Step 1: Interpret the given differential equation

The given differential equation is \(\frac{y_{2}}{y_{1}} = 1 \). This implies that the second derivative \( y_{2} \) equals the first derivative \( y_{1} \).

Step 2: Solve the differential equation

Since \( y_2 = y_1 \), we have the differential equation \( \frac{d^2 y}{dx^2} = \frac{dy}{dx} \), which can be rewritten as \( y'' = y' \). This means the rate of change of the slope (the second derivative) is equal to the slope (the first derivative).

Step 3: Integrate to find the function form

To solve \( y'' = y' \), we first integrate both sides with respect to \( x \) to obtain \( y' = Ce^x \), where \( C \) is a constant.

Step 4: Integrate again to find the general solution

Integrating \( y' = Ce^x \) with respect to \( x \) gives \( y = Ce^x + D \), where \( D \) is another constant of integration.

Step 5: Apply the initial condition

The function passes through \( (0,0) \). Substituting \( x = 0 \) and \( y = 0 \) into the equation \( y = Ce^x + D \), we get \( 0 = Ce^0 + D \), which simplifies to \( 0 = C + D \). This implies \( D = -C \).

Step 6: Simplify the solution with the initial condition

Substituting \( D = -C \) into \( y = Ce^x + D \), we have \( y = Ce^x - C = C(e^x - 1) \). This matches option (D) \( y = k(e^x - 1) \), where \( k \) is a constant.

Key Concepts

Second Derivative

In differential equations, understanding the concept of derivatives is crucial, especially the second derivative. The second derivative of a function indicates how the rate of change of a quantity itself changes. It is essentially the derivative of the first derivative.
Think of the second derivative as the acceleration if the first derivative is speed. If you are driving a car, the speedometer shows your first derivative— your speed at any given moment. The second derivative shows whether your speed is increasing (positive second derivative) or decreasing (negativesecond derivative).
In the context of this problem,

• the differential equation offers that the second derivative, denoted by \(y_2\) or \(\frac{d^2y}{dx^2}\), is equal to the first derivative \(y_1\) or \(\frac{dy}{dx}\).
• This creates an interesting scenario where the rate of change of the rate of change is exactly equal to the initial rate of change.

Integration

Differential equations often require integration to solve. Integration is the process of finding a function when its derivative is known.
It is the reverse process of differentiation.
To solve the equation \(y'' = y'\), we integrate both sides.

• First, integrating both sides with respect to \(x\), we solve for the first derivative: \(y' = Ce^x\), where \(C\) is an integration constant.
• Next, we integrate \(y' = Ce^x\) again to find \(y\), the original function, resulting in \(y = Ce^x + D\), where \(D\) is another constant of integration.
• These constants \(C\) and \(D\) allow for the adjustment needed to fit specific conditions of particular solutions, such as initial conditions.

Initial Conditions

Initial conditions are specific values that a solution to a differential equation must satisfy. They are crucial for determining the particular solution among the infinite number of possibilities provided by the general solution.
In this exercise, the function is supposed to pass through the point \((0,0)\).

• Using the initial condition \(y(0) = 0\), we substitute \(x = 0\) and \(y = 0\) into the general solution \(y = Ce^x + D\).
• This results in \(0 = Ce^0 + D\), which simplifies to \(0 = C + D\). Solving this equation gives \(D = -C\).
• Substituting \(D = -C\) back into the solution transforms it to \(y = C(e^x - 1)\), the particular solution that satisfies the initial condition.
• This ties the solution to option (D) in the given problem.