Q.Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation: $y = e^x + 1$ : $y'' - y' = 0$

The key idea is to verify a solution to an Initial Value Problem by substituting the function and its derivatives into the differential equation.

Step 1: Compute the first derivative.

$y = e^x + 1 \implies y' = e^x$

Step 2: Compute the second derivative.

$y'' = e^x$

Step 3: Substitute into the left-hand side of the given equation $y'' - y' = 0$.

$e^x - e^x = 0$

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

Why This Works: The Idea of a Solution

A differential equation is a condition on a function and its derivatives. To check whether a given function is a solution, you don't solve anything — you substitute. You compute the required derivatives of the candidate function, plug them into the equation, and see if the equation holds true for every $x$ in the domain.

This is exactly like checking whether $x = 2$ satisfies $x^2 - 4 = 0$: you substitute and verify. The only difference is that here, the "unknown" is a function, not a number.

The equation $y'' - y' = 0$ is a second-order linear homogeneous ODE. It says: the second derivative of $y$ must equal the first derivative of $y$, for all $x$. So if we can show that $y''$ and $y'$ are the same function, we're done.

Step-by-Step Verification

1. Compute the first derivative.

Given $y = e^x + 1$, differentiate term by term. The derivative of $e^x$ is $e^x$, and the derivative of the constant $1$ is $0$.

So

2. Compute the second derivative.

Differentiate $y' = e^x$ again. The derivative of $e^x$ is still $e^x$.

So

3. Substitute into the differential equation.

The equation is $y'' - y' = 0$. Replace $y''$ and $y'$ with what we found:

4. Simplify.

$e^x - e^x = 0$ for every real number $x$. The equation holds identically.

5. State the conclusion.

Since the left-hand side equals the right-hand side for all $x$, the function $y = e^x + 1$ is indeed a solution of $y'' - y' = 0$.

Method: Verify a proposed solution by substitution

Use this whenever you are handed a function and a differential equation and asked to confirm the function solves it.

Steps

Step 1: Differentiate the given function as many times as the equation needs.

For a second-order equation, compute both $y'$ and $y''$. Differentiate every term — a constant added to the function differentiates to $0$.

Step 2: Substitute the derivatives into the equation.

Replace $y$, $y'$, $y''$ by the expressions you computed.

Step 3: Simplify and check for an identity.

If the left side reduces to the right side for all $x$, the function is a solution. You are only checking a given function — you never solve the equation from scratch.

Common Mistakes

Mistake 1: Differentiating the constant incorrectly.

Why it's wrong: for $y=e^x+1$, the $+1$ differentiates to $0$, so $y'=e^x$ and $y''=e^x$. Mistakenly writing $y'=e^x+1$ makes $y''-y'=-1\neq 0$ and wrongly rejects a valid solution. Correct approach: the derivative of a constant is $0$.

Mistake 2: Testing at one point instead of as an identity.

Why it's wrong: verification requires the equation to hold for all $x$, not just one value. Correct approach: confirm $e^x-e^x=0$ identically.