Q.Verify that the given implicit function is a solution of the corresponding differential equation: $xy = \log y + C$ ; $y' = \frac{y^2}{1-xy}$ ($xy \neq 1$)

The key idea is implicit differentiation: differentiate both sides of the given relation with respect to $x$, treating $y$ as a function of $x$, and then solve for $y'$ to see if it matches the given differential equation.

Step 1: Differentiate $xy = \log y + C$ implicitly with respect to $x$.

Using the product rule on the left and the chain rule on the right:

Step 2: Collect all terms containing $y'$ on one side.

Factor out $y'$:

Step 3: Solve for $y'$ by isolating it.

$y' \left( \frac{xy - 1}{y} \right) = -y$ …

The core idea here is implicit differentiation. When a relation between $x$ and $y$ is given implicitly (not solved for $y$), we can still find $y'$ by differentiating every term with respect to $x$, treating $y$ as a function of $x$. This means whenever we differentiate a term containing $y$, we multiply by $y'$ (by the chain rule). The goal is to see if the derivative we obtain matches the given $y'$.

Let’s walk through it.

1. Start with the given implicit function:

Here, $C$ is a constant. We need to show that this relation implies the differential equation $y' = \frac{y^2}{1-xy}$.

2. Differentiate both sides with respect to $x$:
   • On the left, $xy$ is a product. Using the product rule: derivative of $x$ is $1$, so we get $1 \cdot y + x \cdot y' = y + x y'$.
   • On the right, $\log y$ differentiates to $\frac{1}{y} \cdot y'$ (chain rule). The constant $C$ differentiates to $0$. So we have:

3. Collect the $y'$ terms on one side: Bring $x y'$ to the right side (or equivalently, move $\frac{y'}{y}$ to the left):

Factor out $y'$ from the right-hand side:

4. Simplify the bracket: Write $\frac{1}{y} - x$ as a single fraction:

So the equation becomes:

5. Solve for $y'$: Multiply both sides by $y$:

Now divide by $(1 - xy)$ (and note the condition $xy \neq 1$ ensures this is safe): …

Method: Verify an implicit solution by implicit differentiation

Use this when the solution is given as a relation like $xy=\log y+C$ that cannot be solved for $y$ explicitly.

Steps

Step 1: Differentiate every term with respect to $x$, treating $y=y(x)$.

Use the product rule on $xy$ (giving $y+xy'$) and the chain rule on $\log y$ (giving $\frac{y'}{y}$). The constant differentiates to $0$.

Step 2: Collect the $y'$ terms and solve for $y'$. …

Common Mistakes

Mistake 1: Forgetting the chain rule on $\log y$.

Why it's wrong: since $y=y(x)$, $\frac{d}{dx}(\log y)=\frac{y'}{y}$, not $\frac{1}{y}$. Dropping the $y'$ makes it impossible to solve for the derivative. Correct approach: differentiate implicitly, keeping the $y'$.

Mistake 2: Sign slip converting $xy-1$ to $1-xy$. …