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

The key idea is Verification of Solution: substitute the given function into the differential equation and check that it holds identically.

Step 1: Differentiate $y = \cos x + C$ with respect to $x$:

Step 2: Substitute $y'$ into the differential equation $y' + \sin x = 0$:

Step 3: The left-hand side simplifies to $0$, which equals the right-hand side for all $x$.

Why This Approach Works

When we say a function is a "solution" of a differential equation, we mean that plugging the function (and its derivatives) into the equation makes it true for every $x$ in the domain. This is exactly like checking whether a number satisfies an algebraic equation — you substitute and simplify. The only difference is that here, the "unknown" is a whole function, and the equation involves its derivative.

The given function $y = \cos x + C$ contains an arbitrary constant $C$. That's expected: the differential equation $y' + \sin x = 0$ is first-order, so its general solution should have one constant. Our job is to confirm that any choice of $C$ works — meaning the family of functions $y = \cos x + C$ truly satisfies the equation.

Step-by-Step Verification

1. Differentiate the candidate function.

We have $y = \cos x + C$. Since $C$ is a constant, its derivative is zero. The derivative of $\cos x$ is $-\sin x$. So:

2. Substitute into the differential equation.

The equation we need to check is $y' + \sin x = 0$. Replace $y'$ with $-\sin x$:

3. Simplify.

$-\sin x + \sin x = 0$ for every real $x$. The equation holds identically — no restrictions, no special cases.

4. Interpret the result.

Because the equality is true for all $x$, the function $y = \cos x + C$ (with any constant $C$) is indeed a solution of $y' + \sin x = 0$. The constant $C$ disappears during differentiation, so it doesn't affect the verification.

Method: Verify a solution involving trigonometric terms

Use this for candidates like $y=\cos x+C$ against $y'+\sin x=0$.

Steps

Step 1: Differentiate carefully, respecting trig signs.

$\frac{d}{dx}(\cos x)=-\sin x$ and $\frac{d}{dx}(C)=0$, so $y'=-\sin x$. A sign slip here is the usual failure point.

Step 2: Substitute into the equation.

Replace $y'$ in $y'+\sin x$ to get $-\sin x+\sin x$.

Step 3: Check the identity.

It simplifies to $0$ for all $x$ and any $C$, so the function is a solution. Verification never requires isolating $y$ or solving the equation.

Common Mistakes

Mistake 1: Sign error differentiating $\cos x$.

Why it's wrong: $\frac{d}{dx}(\cos x)=-\sin x$, so $y'=-\sin x$ and $y'+\sin x=0$. Writing $+\sin x$ for the derivative makes the check fail spuriously. Correct approach: differentiate $\cos x$ to $-\sin x$.

Mistake 2: Treating $C$ as variable.

Why it's wrong: $C$ is a constant, so its derivative is $0$ and it plays no role in the check. Correct approach: the verification holds for every constant $C$.