Key Concepts and Formulas

• Homogeneous Differential Equation: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ where $f(tx, ty) = f(x,y)$ for all $t$. This can be rewritten as $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$.
• Substitution for Homogeneous Equations: Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Separation of Variables: A technique to solve differential equations by isolating variables on each side of the equation before integrating.

Step-by-Step Solution

Step 1: Identify the Differential Equation Type

The given differential equation is: $\frac{dy}{dx} = \frac{x + y}{x}$ Our goal is to show that this is a homogeneous differential equation.

Why? Identifying the equation type allows us to choose the appropriate solution method.

We rewrite the right-hand side: $\frac{dy}{dx} = \frac{x}{x} + \frac{y}{x} = 1 + \frac{y}{x}$ The equation is now in the form $\frac{dy}{dx} = g\left(\frac{y}{x}\right)$, where $g(u) = 1 + u$. Thus, the equation is homogeneous.

Step 2: Apply the Substitution

Let $y = vx$. Then, we need to find $\frac{dy}{dx}$.

Why? This substitution transforms the homogeneous equation into a separable equation.

Differentiating $y = vx$ with respect to $x$ using the product rule gives: $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Step 3: Substitute and Simplify

Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$ into the differential equation: $v + x\frac{dv}{dx} = 1 + \frac{vx}{x}$

Why? This step replaces $y$ and $\frac{dy}{dx}$ with expressions involving $v$ and $x$.

Simplify the equation: $v + x\frac{dv}{dx} = 1 + v$ $x\frac{dv}{dx} = 1$

Step 4: Separate Variables and Integrate

Separate the variables: $dv = \frac{dx}{x}$

Why? Separation allows us to integrate each side independently.

Integrate both sides: $\int dv = \int \frac{dx}{x}$ $v = \ln|x| + C$

Since we are given the initial condition $y(1) = 1$, we know that $x > 0$, so we can drop the absolute value: $v = \ln x + C$

Step 5: Back-Substitute

Substitute $v = \frac{y}{x}$ back into the equation: $\frac{y}{x} = \ln x + C$

Why? We need to express the solution in terms of the original variables $x$ and $y$.

Solve for $y$: $y = x\ln x + Cx$

Step 6: Apply Initial Condition

Apply the initial condition $y(1) = 1$: $1 = (1)\ln(1) + C(1)$

Why? The initial condition allows us to find a particular solution by determining the value of $C$.

Since $\ln(1) = 0$: $1 = 0 + C$ $C = 1$

Step 7: Write the Particular Solution

Substitute $C = 1$ back into the general solution: $y = x\ln x + x$

Step 8: Verify the Solution We have $y = x\ln x + x$. Then $\frac{dy}{dx} = \ln x + x\cdot \frac{1}{x} + 1 = \ln x + 2$. Substituting into the original equation, the RHS is $\frac{x+y}{x} = \frac{x + x\ln x + x}{x} = \frac{2x + x\ln x}{x} = 2 + \ln x$. Since the LHS = RHS, the solution is correct.

Common Mistakes & Tips

• Remember to add the constant of integration after integrating.
• Always back-substitute to express the solution in terms of the original variables.
• Carefully apply the initial condition to solve for the constant of integration.

Summary

We solved the homogeneous differential equation by using the substitution $y=vx$. This transformed the equation into a separable form, which was then integrated. Finally, we applied the initial condition to find the particular solution.

The final answer is \boxed{y = x\ln x + x}, which corresponds to option (D).