Question

Given a differential equation \(\dfrac{d^2x}{dt^2} + x= 0\) with \(x(0) \neq 0\).
Which of the following statements is/are true?

• A. \(|x(t)|^2 + \left|\dfrac{dx(t)}{dt}\right|^2= 0\) for all \(t\geq0\)
• B. \(|x(t)|^2 + \left|\dfrac{dx(t)}{dt}\right|^2= c\) for all \(t\geq0\), for some real constant \(c>0\)
• C. \(|x(t)|^2 + \left|\dfrac{dx(t)}{dt}\right|^2= e^{jt}\) for all \(t\geq0\)
• D. \(|x(t)|^2 + \left|\dfrac{dx(t)}{dt}\right|^2= \sin t\) for all \(t\geq0\)

Hint:

Solve as \(x(t)=C_1\cos t + C_2\sin t\), or note \(\frac{d}{dt}(x^2+(x')^2)=2x'(x+x'')=0\) since \(x''=-x\); the sum is a constant fixed by \(x(0)\neq0\).

Answer 1

The Correct Option is B

Step 1: Solve the differential equation.
The equation
\[ \frac{d^2x}{dt^2}+x=0 \]
is the standard simple harmonic oscillator equation. Its characteristic roots are \(\pm j\), so the general real solution is
\[ x(t) = C_1\cos t + C_2\sin t \]
where \(C_1,C_2\) are real constants set by the initial conditions. Differentiating,
\[ \frac{dx}{dt} = -C_1\sin t + C_2\cos t \]
Step 2: Use the given initial condition to pin down C1.
At \(t=0\), \(x(0)=C_1\cos0+C_2\sin0=C_1\). We are told \(x(0)\neq0\), so
\[ C_1\neq0 \]
Step 3: Compute the quantity asked about, \(|x(t)|^2+|dx/dt|^2\).
\[ |x(t)|^2 = (C_1\cos t+C_2\sin t)^2 = C_1^2\cos^2t + 2C_1C_2\cos t\sin t + C_2^2\sin^2t \]
\[ \left|\frac{dx}{dt}\right|^2 = (-C_1\sin t+C_2\cos t)^2 = C_1^2\sin^2t - 2C_1C_2\sin t\cos t + C_2^2\cos^2t \]
Step 4: Add the two expressions and simplify.
Adding term by term, the cross terms \(2C_1C_2\cos t\sin t\) and \(-2C_1C_2\sin t\cos t\) cancel exactly, and the \(\cos^2t+\sin^2t=1\) identity collects the rest:
\[ |x(t)|^2+\left|\frac{dx}{dt}\right|^2 = C_1^2(\cos^2t+\sin^2t) + C_2^2(\sin^2t+\cos^2t) = C_1^2+C_2^2 \]
This does not depend on \(t\) at all: it is a fixed real number, call it \(c=C_1^2+C_2^2\), true for every \(t\geq0\).
Step 5: Show this constant is strictly positive.
Since \(C_1\neq0\) (from Step 2), \(C_1^2>0\), and \(C_2^2\geq0\) always. So
\[ c = C_1^2+C_2^2 \geq C_1^2 > 0 \]
This is exactly what option (B) states: a real constant \(c>0\), same for all \(t\geq0\). Option (B) is TRUE.
Step 6: Rule out option (A).
Option (A) says the quantity equals \(0\) for all \(t\). We just showed it equals \(c>0\), never \(0\), because \(x(0)\neq0\) forces \(C_1\neq0\). So (A) is FALSE.
Step 7: Rule out option (C).
Option (C) says the quantity equals \(e^{jt}\), which is a complex number that spins around the unit circle as \(t\) changes (its real and imaginary parts are \(\cos t\) and \(\sin t\)). But \(|x(t)|^2+|dx/dt|^2\) is a sum of two squares of real numbers, so it is always a fixed non-negative real number, never complex and never changing with \(t\). At \(t=0\) alone, \(e^{j0}=1\) is real, but the moment \(t\) increases even slightly, \(e^{jt}\) picks up a non-zero imaginary part while our constant \(c\) stays completely real, so equality cannot hold for all \(t\geq0\). Option (C) is FALSE.
Step 8: Rule out option (D).
Option (D) says the quantity equals \(\sin t\), which oscillates between \(-1\) and \(1\) and is even negative for \(t\) in ranges like \((\pi,2\pi)\). Our quantity is a constant \(c>0\) for every \(t\), so it can never trace out \(\sin t\); in particular \(\sin t\) becomes negative while a sum of two real squares never can. Option (D) is FALSE.
Final Answer:
Only option (B) is true:
\[ \boxed{|x(t)|^2+\left|\tfrac{dx(t)}{dt}\right|^2=c,\ c>0,\ \text{constant for all }t\geq0} \]