Problem 109 - A Complete Resource Book in Mathematics for JEE Main

Question

The solution of the differential equation \(\frac{d y}{d x}=\frac{x+y}{x}\) satisfying the condition \(y(1)=1\) is \([2008]\) (A) \(y=\ln x+x\) (B) \(y=x \ln x+x^{2}\) (C) \(y=x e^{(x-1)}\) (D) \(y=x \ln x+x\)

Step-by-Step Solution

Answer

(D) \(y = x \ln x + x\)

Step 1: Simplify the Differential Equation

Start with the given differential equation \( \frac{dy}{dx} = \frac{x+y}{x} \). This can be rewritten as \( \frac{dy}{dx} = 1 + \frac{y}{x} \). The equation is separable, so we can rearrange and proceed to integrate.

Step 2: Separate Variables and Integrate

Separate variables to get \( \frac{dy}{dx} - \frac{y}{x} = 1 \). This is a linear differential equation which can be written in standard form as \( \frac{dy}{dx} - \frac{y}{x} = 1 \). The integrating factor is \( e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \). Multiply the entire equation by this integrating factor to obtain \( \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = \frac{1}{x} \). Integrating both sides gives: \( y = x \ln x + Cx \).

Step 3: Apply Initial Condition

Use the initial condition \( y(1) = 1 \) to solve for the constant \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the integrated equation: \( 1 = 1 \ln 1 + C \cdot 1 \), which simplifies to \( 1 = C \). Thus, \( C = 1 \).

Step 4: Final Solution

Substitute the value of \( C \) back into the integrated equation to get the final solution: \( y = x \ln x + x \).

Step 5: Verify the Solution

Check that the solution satisfies the original differential equation and initial condition. Substituting \( y = x \ln x + x \) back into the differential equation should satisfy it, ensuring the validity of our solution.

Key Concepts

Separable Differential Equations

Separable differential equations are a fundamental class of equations in differential calculus. They are called 'separable' because the variables can be separated into two sides of the equation, allowing us to integrate each side independently. Consider the basic form: \[ \frac{dy}{dx} = f(x)g(y) \] To solve these, rearrange the terms to isolate each variable: \[ \frac{1}{g(y)}dy = f(x)dx \] Now, we can integrate both sides separately. This is the beauty of separable equations - once variables are separated, solving involves simple integration. This approach works when both sides can be expressed in terms of independent variables. Remember that constants of integration appear when you solve these integrals. Applying initial conditions later helps determine these constants. Separate variables if possible, integrate, and use initial conditions to find specific solutions.

Integrating Factor Method

The Integrating Factor Method is often used for solving linear differential equations of the form: \[ \frac{dy}{dx} + P(x)y = Q(x) \] Here's where the magic happens. You introduce a function, known as the integrating factor \( \mu(x) \), which simplifies the solution process. For standard linear equations, the integrating factor is: \[ \mu(x) = e^{\int P(x)dx} \] Multiplying the entire differential equation by \( \mu(x) \) allows the equation to be rewritten: \[ \mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x) \] This step transforms the left side into the derivative of a product, namely \( \frac{d}{dx}[\mu(x)y] \). Integrating both sides with respect to \(x\) yields a solution for \( y \). This method reduces the complexity and provides a straightforward path to the solution.

Initial Value Problems

Initial Value Problems (IVPs) involve finding a particular solution to a differential equation that satisfies a given condition at a specific point. These problems require both solving the differential equation and applying initial conditions to determine any constants in the solution. Consider an IVP given by: \[ \frac{dy}{dx} = f(x,y), \quad y(x_0) = y_0 \] The condition \( y(x_0) = y_0 \) is crucial. It helps to refine the general solution of the differential equation to a specific one that passes through the point \( (x_0, y_0) \). After integrating and finding a general solution with an arbitrary constant, substitute \( x_0 \) and \( y_0 \) into this solution. This step allows you to solve for the arbitrary constant. The determined value ensures the solution aligns with the initial condition. In summary, the IVP framework provides a complete solution customized to start at a specified point.