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

Concept: Verification of Solution — substitute the function into the differential equation and check that it satisfies the equation identically.

Step 1: Differentiate the given function.

$y = x^2 + 2x + C \quad\Rightarrow\quad y' = 2x + 2$

Step 2: Substitute $y'$ into the differential equation $y' - 2x - 2 = 0$:

$(2x + 2) - 2x - 2 = 0$

Step 3: Simplify:

$0 = 0$, which is true for all $x$ and any constant $C$.

Why This Approach Works

When we say a function is a solution to a differential equation, we mean that plugging the function (and its derivatives) into the equation makes it true for all values of the independent variable. Here, the differential equation is first-order — it involves only $y'$ and $x$. So the verification is straightforward: compute $y'$ from the given $y$, substitute into $y' - 2x - 2$, and check if the result is identically zero.

The constant $C$ is a parameter from integration; if the equation holds regardless of $C$, then the whole family of curves is a solution.

Step-by-Step Verification

1. Differentiate the given function We have $y = x^2 + 2x + C$. Differentiating term by term with respect to $x$:

2. Substitute into the differential equation The equation is $y' - 2x - 2 = 0$. Replace $y'$ with $2x + 2$:

3. Simplify

The left-hand side equals the right-hand side (0) for every real $x$.

4. Interpret the result Since the substitution yields $0 = 0$ identically, the function $y = x^2 + 2x + C$ satisfies the differential equation $y' - 2x - 2 = 0$ for any constant $C$. This is expected because the differential equation is first-order and the solution contains one arbitrary constant.

Method: Verify a solution that contains an arbitrary constant

Use this when the candidate carries a constant $C$, e.g. $y=x^2+2x+C$ against $y'-2x-2=0$.

Steps

Step 1: Differentiate; the constant drops out.

$\frac{d}{dx}(C)=0$, so $y'$ carries no $C$. This is expected — a first-order equation has a one-constant family of solutions.

Step 2: Substitute into the equation.

Plug $y'$ into the left side and simplify.

Step 3: Confirm it holds for every $C$.

If the expression reduces to $0=0$ independently of $C$, the whole family is verified. A sanity check: integrating the equation should return the given function, constant and all.

Common Mistakes

Mistake 1: Trying to solve for the constant $C$.

Why it's wrong: $C$ differentiates to $0$, so $y'=2x+2$ regardless of $C$. Attempting to "find $C$" misunderstands the task — the whole family is a solution. Correct approach: verify the identity holds for every $C$.

Mistake 2: Concluding it works only for a specific $C$.

Why it's wrong: substituting gives $0=0$ independent of $C$. Correct approach: a first-order equation has a one-constant family, all of which satisfy it.