Problem 38 - 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 (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

The solution is (D) \( y = x \ln x + x \).

Step 1: Rewrite the Differential Equation

The differential equation given is \( \frac{d y}{d x} = \frac{x+y}{x} \). You can simplify the equation by splitting the fraction to get \( \frac{d y}{d x} = 1 + \frac{y}{x} \).

Step 2: Use Substitution to Solve

Using the substitution \( v = \frac{y}{x} \), we have \( y = vx \) and \( \frac{d y}{d x} = v + x \frac{dv}{dx} \). Substitute these into the differential equation: \( v + x \frac{dv}{dx} = 1 + v \). Simplifying this, we get \( x \frac{dv}{dx} = 1 \).

Step 3: Integrate for v

From the equation \( \frac{dv}{dx} = \frac{1}{x} \), integrate with respect to \( x \). This gives \( v = \ln x + C \), where \( C \) is a constant of integration.

Step 4: Express y in Terms of x

Recall that \( v = \ln x + C = \frac{y}{x} \). Therefore, \( y = x(\ln x + C) \). Simplifying, you get \( y = x \ln x + Cx \).

Step 5: Apply Initial Condition

Use the given initial condition \( y(1) = 1 \) to find \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the equation: \( 1 = 1 \ln 1 + C \cdot 1 \), which simplifies to \( 1 = C \).

Step 6: Write the Final Solution

Substitute \( C = 1 \) back into the equation \( y = x \ln x + Cx \) to get the final solution: \( y = x \ln x + x \). Compare with the options, the solution matches option \( D \).

Key Concepts

Substitution Method

When solving differential equations, the substitution method is a powerful tool to simplify the problem. This method involves replacing a variable or altering an equation format to make integration or further calculation easier. In our given exercise, the complex part of the differential equation is the term \( \frac{y}{x} \). We use the substitution \( v = \frac{y}{x} \), allowing us to express \( y \) as \( vx \). This substitution transforms the original differential equation into a simpler form.

• The goal of substitution is to reduce the equation to a variable we can easily integrate.
• In complex fractions, substitution often involves setting a troublesome term as a new variable.
• This method often simplifies initial differential equations to separable forms.

By changing the form of the equation, we can eliminate the non-linear component temporarily, integrating it more easily later on.

Integration

Integration is a fundamental component in solving differential equations. Once we have simplified our equation using substitution, our task is to integrate. In this scenario, after applying substitution, we ended up with the equation \( \frac{dv}{dx} = \frac{1}{x} \), which is straightforward to integrate.

• The integral of \( \frac{1}{x} \) with respect to \( x \) is \( \ln x \).
• Don't forget the constant of integration, \( C \), which is essential for solving differential equations with given conditions.

After integrating, we get \( v = \ln x + C \), which then enables us to express \( y \) in terms of \( x \), bringing us one step closer to our solution. Integration can thus convert our differential form into an explicit function.

Initial Conditions

Initial conditions are vital to finding a unique solution to a differential equation. They are used to determine the constant of integration, which is often introduced during the integration step. In our specific problem, the initial condition is given by \( y(1) = 1 \).

• Apply initial conditions after solving the differential equation to the general solution.
• Plug the values of the condition into your general solution to solve for the constant \( C \).

For this exercise, inserting \( x = 1 \) and \( y = 1 \) into \( y = x \ln x + Cx \), we solve for \( C \). We find that \( C = 1 \), providing us the unique solution: \( y = x \ln x + x \). Using initial conditions ensures the solution fits the specific scenario described.