Q.(viii) Differential equation representing the family of curves $y=e^x(A\cos x+B\sin x)$ is $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$. (State True or False.)

Concept: Verification of Solution — We check whether the given function satisfies the differential equation by computing its derivatives and substituting.

Step 1: Given $y = e^x (A\cos x + B\sin x)$. Differentiate using the product rule:

Step 2: Differentiate again:

But $e^x(-A\sin x + B\cos x) = \frac{dy}{dx} - y$, and $e^x(-A\cos x - B\sin x) = -y$. So: …

The core idea here is verification of a solution. When a differential equation is claimed to represent a family of curves, we need to check whether every curve in that family actually satisfies the equation. The family has two arbitrary constants $A$ and $B$, so we expect a second-order differential equation — and we must eliminate those constants by differentiating.

Let’s see why this works. The given function is a product of $e^x$ and a linear combination of $\cos x$ and $\sin x$. Its derivatives will involve both the exponential growth and the oscillatory parts. The differential equation $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$ is actually the characteristic equation $r^2-2r+2=0$ whose roots are $1\pm i$ — exactly the exponents in $e^x\cos x$ and $e^x\sin x$. So the family is the general solution of that equation. That’s the conceptual reason the statement is true.

Now let’s verify step by step.

1. Start with the given family.

2. First derivative. Use the product rule: differentiate $e^x$ times the bracket, plus $e^x$ times the derivative of the bracket.

Factor $e^x$:

Group like terms:

3. Second derivative. Differentiate $\frac{dy}{dx}$ again. It’s still $e^x$ times a linear combination of $\cos x$ and $\sin x$, so apply the product rule similarly. Let $P = (A+B)\cos x + (B-A)\sin x$. Then $\frac{dy}{dx} = e^x P$, so

Compute $\frac{dP}{dx}$:

Therefore:

Combine the $\cos x$ terms: $(A+B) + (B-A) = 2B$

Combine the $\sin x$ terms: $(B-A) - (A+B) = -2A$

So:

4. Form the left-hand side of the given differential equation.

   We need $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y$. Substitute each piece:

   • $y = e^x (A\cos x + B\sin x)$
   • $\frac{dy}{dx} = e^x \big[ (A+B)\cos x + (B-A)\sin x \big]$
   • $\frac{d^2y}{dx^2} = e^x (2B\cos x - 2A\sin x)$

   Compute term by term: …

Method: Forming a Second-Order DE by Eliminating Two Constants

A family with two arbitrary constants ($A$ and $B$) needs a second-order differential equation. Eliminate the constants by differentiating twice and expressing higher derivatives through $y$ and $\frac{dy}{dx}$.

Steps

Step 1: Differentiate the family once.

Differentiate $y$ (using the product rule for terms like $e^{x}(A\cos x+B\sin x)$); write the result so part of it is recognisably $y$ itself.

Step 2: Differentiate again and re-express.

Compute $\frac{d^2y}{dx^2}$ and substitute back the combinations already identified as $y$ and $\frac{dy}{dx}$, so $A$ and $B$ cancel. …

Common Mistakes

Mistake 1: Solving for $A$ and $B$ individually instead of eliminating them.

Why it's wrong: forming the differential equation means removing both constants, not finding them; you re-express $\frac{d^2y}{dx^2}$ using $y$ and $\frac{dy}{dx}$ so $A,B$ cancel. Correct approach: differentiate twice and substitute back the identified combinations.

Mistake 2: Reaching for the auxiliary/characteristic-equation shortcut. …