In each of the Exercises 1 to 10, show that the given differential…

Question 48 of 96intermediate🔧 ApplyLong Answer4 marks

In each of the Exercises 1 to 10, show that the given differential equation is homogeneous and solve each of them.

y' = \frac{x + y}{x}

Correct Answer

The general solution is $y = x \log|x| + Cx$ , where $C$ is an arbitrary constant.

Exercise: EXERCISE 9.4 | Q: 2 | (Page 22)

For More Understanding

Explanation

This question is from Exercise 9.4, Question 2. A homogeneous differential equation can be expressed in the form $\frac{dy}{dx} = f\!\left(\frac{y}{x}\right)$ .

The given equation simplifies to $\frac{dy}{dx} = 1 + \frac{y}{x}$ , which is clearly a function of $\frac{y}{x}$ , proving homogeneity.

The standard method for solving homogeneous equations involves the substitution $y = vx$ , which transforms the equation into a separable form in variables $v$ and $x$ .

Solution Steps

Step 1: Rewrite the equation as $\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$ . This is of the form $\frac{dy}{dx} = f\!\left(\frac{y}{x}\right)$ , hence the equation is homogeneous.
Step 2: Substitute $y = vx$ , so $\frac{dy}{dx} = v + x\frac{dv}{dx}$ .
Step 3: Substituting into the equation:
$v + x\frac{dv}{dx} = 1 + v$
which simplifies to $x\frac{dv}{dx} = 1$ .
Step 4: Separate variables:
$dv = \frac{1}{x}\,dx$
Step 5: Integrate both sides:
$\int dv = \int \frac{1}{x}\,dx$
giving $v = \log|x| + C$ .
Step 6: Substitute back $v = \frac{y}{x}$ :
$\frac{y}{x} = \log|x| + C$
∴ $y = x \log|x| + Cx$

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