Q.Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $x + y = \tan^{-1} y$ : $y^2 y' + y^2 + 1 = 0$

Concept: Implicit Differentiation — differentiate the given relation with respect to $x$, treating $y$ as a function of $x$, then substitute into the differential equation.

Step 1: Differentiate $x + y = \tan^{-1} y$ with respect to $x$:

Step 2: Multiply through by $1 + y^2$:

$1 + y^2 + y' + y^2 y' = y'$ …

Why implicit differentiation is the natural tool here

The given relation $x + y = \tan^{-1} y$ defines $y$ as an implicit function of $x$ — we cannot easily solve for $y$ in terms of $x$ (and we don't need to). The differential equation involves $y'$, so we differentiate both sides of the relation with respect to $x$, treating $y$ as a function of $x$. This is the standard technique for verifying implicit solutions.

Step-by-step verification

1. Differentiate the given relation implicitly

We start with:

Differentiate both sides with respect to $x$:

The left side gives $1 + y'$. For the right side, recall that $\frac{d}{dy}(\tan^{-1} y) = \frac{1}{1+y^2}$, so by the chain rule:

Thus:

2. Solve for $y'$

Multiply both sides by $1+y^2$:

Expand the left side:

Subtract $y'$ from both sides:

$1 + y^2 + y^2 y' = 0$ …

Method: Verifying an implicit solution of a differential equation

Use this whenever a question gives you a relation between $x$ and $y$ (like $x+y=\tan^{-1}y$) and asks you to check that it satisfies a given differential equation — you are not solving the equation, only confirming a candidate.

Steps

Step 1: Differentiate the given relation implicitly

Differentiate both sides with respect to $x$, remembering that $y$ is a hidden function of $x$. Every $y$-term therefore picks up a factor $y'$ by the chain rule:

Step 2: Collect and solve for $y'$ (or for the target combination)

Rearrange the differentiated equation to express $y'$, or better, to reproduce the exact grouping of terms that appears in the target differential equation. Often you do not need $y'$ alone — you just need to reach the same algebraic form.

Step 3: Match against the given differential equation …

Common Mistakes

Mistake 1: Differentiating $\tan^{-1}y$ as if $y$ were the variable

Why it's wrong: with respect to $x$, the chain rule gives $\dfrac{d}{dx}\tan^{-1}y = \dfrac{1}{1+y^2}\cdot y'$, not $\dfrac{1}{1+y^2}$. Dropping the $y'$ factor loses the very term the differential equation depends on. Correct approach: attach $y'$ to every $y$-term you differentiate.

Mistake 2: Trying to "solve" the relation instead of verifying it …