Question

The solution of the differential equation \( \frac{dy}{dx} = \frac{x + y}{x} \) satisfying \( y(1) = 1 \) is

• A. \( y = \ln x + x \)
• B. \( y = \ln x + x^2 \)
• C. \( y = x e^{x-1} \)
• D. \( y = x \ln x + x \)

Hint:

For first-order linear differential equations, use the method of integrating factors. The integrating factor simplifies the equation, allowing you to integrate and solve for the unknown function.

Answer 1

The Correct Option is D

Step 1: Understanding the differential equation.
The given differential equation is: \[ \frac{dy}{dx} = \frac{x + y}{x} \] This is a first-order linear differential equation. We can rewrite it as: \[ \frac{dy}{dx} = 1 + \frac{y}{x} \] To solve this equation, we recognize that it is in the form of a linear differential equation, and we can use the method of integrating factor to solve it.
Step 2: Transforming the equation.
Rearrange the equation: \[ \frac{dy}{dx} - \frac{y}{x} = 1 \] This is now in the standard form of a linear differential equation: \[ \frac{dy}{dx} + P(x) y = Q(x) \] where \( P(x) = -\frac{1}{x} \) and \( Q(x) = 1 \).
Step 3: Finding the integrating factor.
The integrating factor is given by: \[ I(x) = e^{\int P(x) dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} \]
Step 4: Multiplying by the integrating factor.
Multiply both sides of the differential equation by the integrating factor \( \frac{1}{x} \): \[ \frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = \frac{1}{x} \] This simplifies to: \[ \frac{d}{dx} \left( \frac{y}{x} \right) = \frac{1}{x} \] Now, we integrate both sides with respect to \( x \).
Step 5: Integrating both sides.
Integrating both sides: \[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int \frac{1}{x} dx \] The left side becomes \( \frac{y}{x} \), and the right side integrates to \( \ln x \): \[ \frac{y}{x} = \ln x + C \] where \( C \) is the constant of integration.
Step 6: Solving for \( y \).
Multiply both sides by \( x \) to solve for \( y \): \[ y = x (\ln x + C) \]
Step 7: Applying the initial condition.
We are given the initial condition \( y(1) = 1 \). Substitute \( x = 1 \) and \( y = 1 \) into the equation: \[ 1 = 1 \times (\ln 1 + C) \] Since \( \ln 1 = 0 \), we get: \[ 1 = C \] Thus, \( C = 1 \).
Step 8: Final solution.
Substitute \( C = 1 \) into the general solution: \[ y = x (\ln x + 1) \] Thus, the solution to the differential equation is: \[ y = x \ln x + x \]
Step 9: Conclusion.
The correct solution is (D) \( y = x \ln x + x \).

Answer 2

This equation is homogeneous in \( x \) and \( y \) (the right-hand side depends only on the ratio \( y/x \)), so it can be solved by the standard homogeneous-equation substitution instead of the integrating-factor method.

Step 1: Substitute \( y = vx \).
Then \( \dfrac{dy}{dx} = v + x\dfrac{dv}{dx} \). The equation \( \dfrac{dy}{dx} = \dfrac{x+y}{x} = 1 + \dfrac{y}{x} \) becomes \( v + x\dfrac{dv}{dx} = 1 + v \).

Step 2: Simplify and separate variables.
The \( v \) terms cancel, leaving \( x\dfrac{dv}{dx} = 1 \), so \( dv = \dfrac{dx}{x} \).

Step 3: Integrate both sides.
\( \int dv = \int \dfrac{dx}{x} \) gives \( v = \ln x + C \).

Step 4: Substitute back \( v = y/x \).
This gives \( \dfrac{y}{x} = \ln x + C \), so \( y = x\ln x + Cx \).

Step 5: Apply the initial condition \( y(1) = 1 \).
At \( x = 1 \): \( 1 = 1\cdot(\ln 1 + C) = C \), since \( \ln 1 = 0 \), so \( C = 1 \).

Substituting back gives \( y = x\ln x + x \).

Therefore, the correct answer is \( y = x\ln x + x \).