Q.Verify that given function (explicit or implicit) is a solution of the corresponding differential equation: $y=\sqrt{1+x^2}$ ; $\frac{dy}{dx}=\frac{xy}{1+x^2}$

Solution Verification — The Intuition

When you solve an equation, you perform operations on both sides to isolate the variable. But here's the catch: some operations are reversible (like adding 5 or multiplying by 3) and some are not (like squaring both sides or multiplying by an expression that could be zero). When you use a non-reversible step, you might introduce extra "solutions" that don't actually satisfy the original equation. These are called extraneous solutions.

Solution verification is the simple act of plugging your final answers back into the original equation to check which ones actually work. It's your safety net against these fake solutions.

The Precise Statement

Definition:

Given an equation $E(x) = 0$ and a set of candidate solutions $S = \{x_1, x_2, \dots, x_n\}$ obtained by algebraic manipulation, solution verification is the process of substituting each $x_i$ into the original equation $E(x) = 0$ and retaining only those $x_i$ for which the equality holds true.

Why it's necessary:

If at any step you performed an operation that is not bijective (one-to-one and onto) on the domain of the equation, the solution set of the transformed equation may be a superset of the solution set of the original equation. Common culprits:

• Squaring both sides: $x = 2 \implies x^2 = 4$, but $x^2 = 4$ also gives $x = -2$, which is extraneous.
• Multiplying by an expression that could be zero: $\frac{x}{x-1} = 2 \implies x = 2(x-1)$ loses the restriction $x \neq 1$.
• Taking logarithms or exponentials without domain checks.

A Worked Example

Solve: $\sqrt{x+6} = x$

Step 1 — Square both sides:

$(\sqrt{x+6})^2 = x^2 \implies x+6 = x^2 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0$

So $x = 3$ or $x = -2$.

Step 2 — Verify in the original equation:

• For $x = 3$: $\sqrt{3+6} = \sqrt{9} = 3$, and RHS is $3$. Works.
• For $x = -2$: $\sqrt{-2+6} = \sqrt{4} = 2$, but RHS is $-2$. $2 \neq -2$. Extraneous.

Final answer: $x = 3$ only.

When You Can Skip Verification

You do not need to verify if every step you performed was a reversible transformation on the entire domain of the original equation. These include:

• Adding or subtracting any expression (constant or variable)
• Multiplying or dividing by a non-zero constant
• Applying a strictly monotonic function (like $e^x$, $\log x$, or an odd power like $x^3$)

But in practice, for Indian exams (JEE, board exams), always verify unless the problem explicitly says "without checking" or the solution is trivial. It costs 10 seconds and saves marks.