Question

The general solution of the differential equation \( y\,dx-x\,dy=y^2(x\,dy+y\,dx) \) is

• A. \( \dfrac{y}{x}=xy+C \)
• B. \( \dfrac{x}{y^2}=xy+C \)
• C. \( \dfrac{y}{x^2}=xy+C \)
• D. \( x+y=xy+C \)
• E. \( \dfrac{x}{y}=xy+C \)

Hint:

In differential equations involving \( dx \) and \( dy \), first collect like terms carefully. After separation, partial fractions often make the integration straightforward.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
The given differential equation is not in a standard form. We need to rearrange it and identify exact differentials to solve it. The terms \( ydx - xdy \) and \( xdy + ydx \) are standard differential forms.
Step 2: Key Formula or Approach:
Recognize the following exact differentials:
1. \( d(xy) = xdy + ydx \)
2. \( d\left(\frac{x}{y}\right) = \frac{y dx - x dy}{y^2} \)
3. \( d\left(\frac{y}{x}\right) = \frac{x dy - y dx}{x^2} \)
We will manipulate the given equation to use these forms.
Step 3: Detailed Explanation:
The given equation is:
\[ ydx - xdy = y^2(xdy + ydx) \] We can immediately recognize the term in the parenthesis on the right side as the differential of a product.
\[ xdy + ydx = d(xy) \] Substituting this into the equation gives:
\[ ydx - xdy = y^2 d(xy) \] Now, look at the left side, \( ydx - xdy \). This is the numerator of the differential of a quotient, \( d(x/y) \). The denominator for \( d(x/y) \) is \( y^2 \). Conveniently, we have a \( y^2 \) term on the right side. Let's divide the entire equation by \( y^2 \) (assuming \( y \neq 0 \)).
\[ \frac{ydx - xdy}{y^2} = d(xy) \] The left side is now the exact differential of \( (x/y) \).
\[ d\left(\frac{x}{y}\right) = d(xy) \] Now that we have separated the equation into exact differentials, we can integrate both sides:
\[ \int d\left(\frac{x}{y}\right) = \int d(xy) \] Integration of a differential \( d(f) \) is just \( f \).
\[ \frac{x}{y} = xy + C \] where C is the constant of integration.
Step 4: Final Answer:
The general solution of the differential equation is \( \frac{x}{y} = xy + C \). This corresponds to option (E).