Question

If $y = P \cos(ux) + Q \sin(ux)$, where $P, Q,$ and $u$ are constants, show that $\frac{d^2y}{dx^2} + u^2y = 0$.

Hint:

When differentiating sine and cosine functions repeatedly, remember that the signs alternate in pairs: sine becomes cosine, cosine becomes negative sine, negative sine becomes negative cosine, and negative cosine returns to positive sine.

Answer 1

Concept: This problem requires verifying a second-order linear homogeneous differential equation. We will find the first derivative ($\frac{dy}{dx}$) and the second derivative ($\frac{d^2y}{dx^2}$) of the given function with respect to $x$. We will then substitute these derivatives back into the differential expression to verify that it simplifies to zero.

Step 1: Finding the first derivative $\frac{dy}{dx}$.
The given function is: \[ y = P \cos(ux) + Q \sin(ux) \] Differentiating both sides with respect to $x$ using the chain rule ($\frac{d}{dx}[\cos(ux)] = -u\sin(ux)$ and $\frac{d}{dx}[\sin(ux)] = u\cos(ux)$): \[ \frac{dy}{dx} = P \cdot \left(-u \sin(ux)\right) + Q \cdot \left(u \cos(ux)\right) \] \[ \frac{dy}{dx} = -Pu \sin(ux) + Qu \cos(ux) \]

Step 2: Finding the second derivative $\frac{d^2y}{dx^2}$.
Now, differentiate the first derivative expression with respect to $x$ once more to find the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left[-Pu \sin(ux) + Qu \cos(ux)\right] \] Apply the chain rule again to both terms: \[ \frac{d^2y}{dx^2} = -Pu \cdot \left(u \cos(ux)\right) + Qu \cdot \left(-u \sin(ux)\right) \] Multiply the constant parameters together: \[ \frac{d^2y}{dx^2} = -Pu^2 \cos(ux) - Qu^2 \sin(ux) \]

Step 3: Factoring out the common constant parameter and substituting $y$.
Let us factor out the common term $-u^2$ from both expressions on the right side: \[ \frac{d^2y}{dx^2} = -u^2 \left[ P \cos(ux) + Q \sin(ux) \right] \] Notice that the expression inside the brackets, $\left[ P \cos(ux) + Q \sin(ux) \right]$, is exactly equal to our original function $y$. Substituting $y$ back into the equation gives: \[ \frac{d^2y}{dx^2} = -u^2y \]

Step 4: Rearranging the terms into the final requested format.
Move the term $-u^2y$ from the right side of the equation to the left side by adding $u^2y$ to both sides: \[ \frac{d^2y}{dx^2} + u^2y = 0 \] This completes the required mathematical proof. *(Hence Proved)*