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

The key idea is to differentiate the given explicit function and check whether the result matches the proposed differential equation.

Step 1: Write $y = \sqrt{1 + x^2} = (1 + x^2)^{1/2}$.

Step 2: Differentiate using the chain rule:

Step 3: Substitute $y = \sqrt{1 + x^2}$ into the right-hand side of the given DE: …

Why This Works: The Idea of Verification

When a problem asks you to "verify that a given function is a solution" of a differential equation, it means: plug the function (and its derivative) into the equation and check that both sides match identically. You don't solve the differential equation — you just confirm that the candidate works.

The key tool here is differentiation. Since $y$ is given explicitly as a function of $x$, we can compute $y'$ directly, then substitute both $y$ and $y'$ into the right-hand side of the equation and see if we get the same expression.

Step-by-Step Verification

1. Write down the given function and the differential equation.

We have:

and the differential equation:

2. Differentiate $y$ with respect to $x$.

Since $y = (1 + x^2)^{1/2}$, use the chain rule:

3. Express $y'$ in terms of $y$ and $x$.

We know $y = \sqrt{1 + x^2}$, so:

4. Compare with the right-hand side of the differential equation.

The given differential equation says:

But we have $y' = \frac{x}{y}$. Are these the same? Let's check by rewriting $\frac{xy}{1 + x^2}$ using the fact that $y^2 = 1 + x^2$. …

Method: Verify a solution using the relation $y^2$ hides

Use this when the candidate is a root like $y=\sqrt{1+x^2}$ and the target equation contains $1+x^2$.

Steps

Step 1: Differentiate with the chain rule.

Writing $y=(1+x^2)^{1/2}$, $y'=\frac{1}{2}(1+x^2)^{-1/2}\cdot 2x=\frac{x}{\sqrt{1+x^2}}$.

Step 2: Express the target's right side using the candidate. …

Common Mistakes

Mistake 1: Chain-rule slip on $\sqrt{1+x^2}$.

Why it's wrong: $\frac{d}{dx}(1+x^2)^{1/2}=\frac{1}{2}(1+x^2)^{-1/2}\cdot 2x=\frac{x}{\sqrt{1+x^2}}$; forgetting the inner $2x$ or the $\frac{1}{2}$ gives a wrong $y'$. Correct approach: apply the chain rule carefully.

Mistake 2: Fighting the radical instead of using $y^2=1+x^2$. …