Q.If $y=e^{-x}(A\cos x+B\sin x)$, then $y$ is a solution of:
(A) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}=0$
(B) $\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=0$
(C) $\frac{d^2y}{dx^2}+2\frac{dy}{dx}+2y=0$
(D) $\frac{d^2y}{dx^2}+2y=0$

Concept: Verification of Solution — substitute the given function into each differential equation and check which one is identically satisfied.

Given $y = e^{-x}(A\cos x + B\sin x)$, differentiate:

First derivative:

Second derivative (differentiate the first derivative form):

Now test option (C): $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$.

Substitute: …

The core idea here is verification of a solution — we are given a candidate function and need to check which differential equation it satisfies. Instead of solving each ODE from scratch, we can compute the derivatives of $y$ and substitute them into each option. But there's a more elegant way: recognise the form.

The function $y = e^{-x}(A\cos x + B\sin x)$ is the general solution of a second-order linear homogeneous ODE with constant coefficients. The characteristic equation for such an ODE is $r^2 + pr + q = 0$, and the solution form $e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$ corresponds to complex conjugate roots $\alpha \pm i\beta$.

Here, $\alpha = -1$ and $\beta = 1$. So the characteristic roots are $-1 \pm i$. The characteristic equation is therefore:

which simplifies to:

Thus the ODE is $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$.

Let's verify this by direct differentiation as well.

1. First derivative:

   $y = e^{-x}(A\cos x + B\sin x)$

   Using the product rule:

   $\frac{dy}{dx} = -e^{-x}(A\cos x + B\sin x) + e^{-x}(-A\sin x + B\cos x)$

   Factor $e^{-x}$:

   $\frac{dy}{dx} = e^{-x}\left[ -A\cos x - B\sin x - A\sin x + B\cos x \right]$

   Group $\cos x$ and $\sin x$ terms:

   $\frac{dy}{dx} = e^{-x}\left[ (-A + B)\cos x + (-B - A)\sin x \right]$

2. Second derivative:

   Differentiate $\frac{dy}{dx}$ again. Let $P = -A + B$ and $Q = -A - B$, so $\frac{dy}{dx} = e^{-x}(P\cos x + Q\sin x)$.

   Then:

   $\frac{d^2y}{dx^2} = -e^{-x}(P\cos x + Q\sin x) + e^{-x}(-P\sin x + Q\cos x)$

   $= e^{-x}\left[ (-P + Q)\cos x + (-Q - P)\sin x \right]$

   Substitute back $P$ and $Q$:

   $-P + Q = -(-A+B) + (-A-B) = A - B - A - B = -2B$

   $-Q - P = -(-A-B) - (-A+B) = A + B + A - B = 2A$

   So $\frac{d^2y}{dx^2} = e^{-x}(-2B\cos x + 2A\sin x)$

3. Now check option (C): $\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y = 0$

   Compute $2\frac{dy}{dx} = 2e^{-x}\left[ (-A + B)\cos x + (-A - B)\sin x \right]$

   And $2y = 2e^{-x}(A\cos x + B\sin x)$

   Add them: …

Method: Testing Which DE a Given Function Satisfies

To find which differential equation a function solves, compute the needed derivatives and substitute into each candidate, keeping the answer that reduces to $0$ identically.

Steps

Step 1: Differentiate the function twice.

For $y=e^{-x}(A\cos x+B\sin x)$, use the product rule; simplify each derivative by grouping $\cos x$ and $\sin x$ coefficients.

Step 2: Substitute into the candidate equation.

Insert $y$, $\frac{dy}{dx}$, $\frac{d^2y}{dx^2}$ and collect the $\cos x$ and $\sin x$ coefficients separately.

Step 3: Require both coefficients to vanish. …

Common Mistakes

Mistake 1: Guessing the equation from $A\cos x+B\sin x$ while ignoring the $e^{-x}$ factor.

Why it's wrong: the decaying factor $e^{-x}$ shifts the behaviour so the middle coefficient is $+2\frac{dy}{dx}$ (option C), not the $-2\frac{dy}{dx}$ you would get for a growing $e^{x}$ factor. Correct approach: differentiate the full product and test the candidate. …