Problem 35 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\frac{d y}{d x}=y+3>0\) and \(y(0)=2\), then \(y(\ln 2)\) is equal to (A) \(-2\) (B) 7 (C) 5 (D) 13

Step-by-Step Solution

Answer

The value of \( y(\ln 2) \) is 7.

Step 1: Identify the Differential Equation

We are given the differential equation \( \frac{dy}{dx} = y + 3 \). This is a first-order linear differential equation which we need to solve.

Step 2: Solve the Differential Equation

To solve this differential equation, we can use the method of separation of variables or linear differential equation method. If we rearrange, we get \( \frac{dy}{dx} - y = 3 \). This is a linear differential equation.

Step 3: Find the Integrating Factor

For the equation \( \frac{dy}{dx} - y = 3 \), the integrating factor \( \mu(x) \) is given by \( e^{-\int 1 \, dx} = e^{-x} \).

Step 4: Solve with Integrating Factor

Multiplying through by the integrating factor, our equation becomes \( e^{-x} \frac{dy}{dx} - e^{-x} y = 3 e^{-x} \). This simplifies to \( \frac{d}{dx}(e^{-x} y) = 3 e^{-x} \).

Step 5: Integrate Both Sides

Integrate both sides with respect to \( x \):\[ e^{-x} y = \int 3 e^{-x} \, dx = -3 e^{-x} + C \].

Step 6: Solve for y

Substitute back, \( y = -3 + Ce^{x} \).

Step 7: Use Initial Condition

Using the initial condition \( y(0) = 2 \):\(-3 + Ce^{0} = 2 \) which means \( C - 3 = 2 \) leading to \( C = 5 \). Thus, \( y = -3 + 5 e^{x} \).

Step 8: Evaluate y at \( x = \ln 2 \)

Substitute \( x = \ln 2 \) into \( y = -3 + 5 e^{x} \):\( y(\ln 2) = -3 + 5 e^{\ln 2} = -3 + 5 \cdot 2 \).

Step 9: Calculate Final Value

Calculate \( y(\ln 2) = -3 + 10 = 7 \). Thus, the value of \( y(\ln 2) \) is 7.

Key Concepts

first-order linear differential equation

A first-order linear differential equation is a type of equation that involves a first derivative only. It does not contain higher order derivatives like the second or third derivatives. The standard form of a first-order linear differential equation looks like this:\[ \frac{dy}{dx} + P(x)y = Q(x) \]Here, \( P(x) \) and \( Q(x) \) are known functions of \( x \). In our exercise case, the original equation given is \( \frac{dy}{dx} = y + 3 \), which can be rearranged into this standard form: \( \frac{dy}{dx} - y = 3 \). These equations are important because they appear frequently, not only in textbook exercises but also in real-world problems involving rates of change and linear growth or decay.
Understanding that such equations are 'linear' is key because it simplifies solving them. The linearity implies solutions can be manipulated in predictable ways. They can be solved using straightforward methods, such as finding an integrating factor. This predictability is incredibly useful when modeling a wide range of natural phenomena.

integrating factor method

The integrating factor method is a powerful technique used to solve first-order linear differential equations. This method simplifies the equation such that it can be easily integrated. The integrating factor \( \mu(x) \) is calculated using the formula:\[ \mu(x) = e^{\int P(x) \, dx} \]In our exercise, the rearranged equation \( \frac{dy}{dx} - y = 3 \) has \( P(x) = -1 \). Thus, the integrating factor \( \mu(x) \) becomes \( e^{-\int 1 \, dx} = e^{-x} \).Using the integrating factor, the differential equation is transformed into a form that allows direct integration on both sides. This is done by multiplying every term in the equation by the integrating factor. This results in the left side becoming a derivative of a product. In this exercise, the equation modifies to:\[ \frac{d}{dx}(e^{-x} y) = 3 e^{-x} \]
This step is crucial because it makes the equation much simpler and facilitates solving it through integration. Once integrated, the solution can then be found by applying the necessary constants through the initial conditions.

initial condition

An initial condition is a crucial part of solving differential equations, as it gives us enough information to solve for any constant terms that arise. Differential equations usually have more than one solution due to the presence of one or more arbitrary constants resulting from integration.
In our exercise, the initial condition provided is \( y(0) = 2 \). When you substitute \( x = 0 \) and \( y = 2 \) into the integrated equation, it allows you to solve for the constant \( C \). With this condition, the integration constant can be adjusted so that the specific solution fits the conditions given at \( x = 0 \). From our solution:\[ y = -3 + Ce^{x} \]Applying the initial condition, we find \( C \) by solving the equation:\(-3 + Ce^{0} = 2\), leading to \( C = 5 \).The initial condition, therefore, enables us to tailor the general solution of a differential equation to a particular solution which precisely fits the model of the situation in hand, uniquely determining the behavior of the system modeled by the differential equation from a given starting point.