Show that x = sin(kt) + cos(kt) satisfies the differential equation x'' + k^2 x = 0.

Step-by-step solution

1. 1

   Plan the verification. To confirm a proposed solution, differentiate it twice, substitute into the equation, and check that the left side reduces to zero identically — for every $t$, not just at sample points.

2. 2

   Differentiate once, applying the chain rule. The inner function is $kt$, whose derivative is $k$, so each term picks up a factor of $k$:

   $x'=k\cos(kt)-k\sin(kt)$

   Note the sign swap: $\sin$ differentiates to $+\cos$, but $\cos$ differentiates to $-\sin$.

3. 3

   Differentiate again. Each term picks up another factor of $k$, and the signs swap once more:

   $x''=-k^2\sin(kt)-k^2\cos(kt)$

   So two differentiations produce $k^2$ and a net sign reversal — the hallmark of sine and cosine.

4. 4

   Factor out -k^2 and recognise the original function.

   $x''=-k^2\left(\sin(kt)+\cos(kt)\right)=-k^2x$

   since the bracket is exactly $x$.

5. 5

   Substitute into the equation.

   $x''+k^2x=-k^2x+k^2x=0\ \checkmark$

   The identity holds for all $t$ and all $k$, so the function is a genuine solution.

6. 6

   Note what this means physically and structurally. $x''+k^2x=0$ is simple harmonic motion, and $\{\cos(kt),\sin(kt)\}$ spans its full two-dimensional solution space — so $\sin(kt)+\cos(kt)$ is the member with both coefficients equal to $1$. It can be rewritten in amplitude-phase form as $\sqrt2\sin\!\left(kt+\tfrac{\pi}{4}\right)$, revealing amplitude $\sqrt2$ and period $\tfrac{2\pi}{k}$.

7. 7

   Verify numerically. With $k=1.7$, a second-order central difference of step $10^{-4}$ gives $x''$ values matching $-k^2x$ at $t=0.3,\,1.1,\,-2.0$ to four significant figures ✓, and the amplitude-phase rewrite agrees with the original at every test point ✓.