Q.Verify that the function, $y=ke^x-1$ is a solution of the differential equation $\frac{dy}{dx}=y+1$. Also determine the value of the constant $k$ so that the solution curve of the given differential equation passes through the point $(0,1)$.

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. …