Question

Consider the differential equation \(y^{\prime \prime}(t)+k^{2} y(t)=0,\) where \(k\) is a positive real number. a. Verify by substitution that when \(k=1\), a solution of the equation is \(y(t)=C_{1} \sin t+C_{2} \cos t .\) You may assume this function is the general solution. b. Verify by substitution that when \(k=2\), the general solution of the equation is \(y(t)=C_{1} \sin 2 t+C_{2} \cos 2 t\) c. Give the general solution of the equation for arbitrary \(k>0\) and verify your conjecture.

Short answer

Question: Verify the given solutions \(y(t) = C_1 \sin t + C_2 \cos t\) and \(y(t) = C_1 \sin 2t + C_2 \cos 2t\) for the differential equation \(y''(t) + k^2 y(t) = 0\) and find the general solution for arbitrary k > 0. Answer: The given solutions are verified to satisfy the equation for \(k=1\) and \(k=2\). The general solution for arbitrary \(k > 0\) is \(y(t) = C_1 \sin kt + C_2 \cos kt\).

Step-by-step solution

Finding the first and second derivatives of the given functions

The given functions are \(y(t) = C_1 \sin t + C_2 \cos t\) and \(y(t) = C_1 \sin 2t + C_2 \cos 2t\). To verify these functions, we need to find their first and second derivatives with respect to t. a. For \(y(t) = C_1 \sin t + C_2 \cos t\): \(y'(t) = C_1 \cos t - C_2 \sin t\) \(y''(t) = -C_1 \sin t - C_2 \cos t\) b. For \(y(t) = C_1 \sin 2t + C_2 \cos 2t\): \(y'(t) = 2C_1 \cos 2t - 2C_2 \sin 2t\) \(y''(t) = -4C_1 \sin 2t - 4C_2 \cos 2t\)

Verifying the solutions by substitution

Now that we have the first and second derivatives, we will plug these values into the given equation separately for \(k = 1\) and \(k = 2\). a. For \(k = 1\): \(y''(t) + k^2 y(t) = -C_1 \sin t - C_2 \cos t + (C_1 \sin t + C_2 \cos t) = 0\) b. For \(k = 2\): \(y''(t) + k^2 y(t) = -4C_1 \sin 2t - 4C_2 \cos 2t + 4(C_1 \sin 2t + C_2 \cos 2t) = 0\) In both the cases, we can see that the equation is satisfied, confirming that the given functions are indeed solutions.

General solution for arbitrary k > 0

Now, we will find the general solution for arbitrary k > 0: The general form of the equation is \(y''(t) + k^2 y(t) = 0\). First, we write down the general form of the solution for any real k. \(y(t) = C_1 \sin kt + C_2 \cos kt\) Taking its first and second derivatives w.r.t to t: \(y'(t) = k C_1 \cos kt - kC_2 \sin kt\) \(y''(t) = -k^2 C_1 \sin kt - k^2 C_2 \cos kt\) Now, we substitute the results into the given equation: \(y''(t) + k^2 y(t) = -k^2 C_1 \sin kt - k^2 C_2 \cos kt + k^2 (C_1 \sin kt + C_2 \cos kt) = 0\) Thus, the general solution for arbitrary \(k > 0\) is: \(y(t) = C_1 \sin kt + C_2 \cos kt\)

Key concepts

General Solution of Differential Equation

The general solution of a differential equation encapsulates all possible solutions to the equation. It's like finding the universal recipe that can explain every outcome of a particular situation. In the context of second-order linear homogeneous differential equations, the general solution often involves constants, which can be determined by initial conditions or specific requirements. For any positive real number k, the general solution to the equation y''(t) + k^2y(t) = 0 is expressed by combining sine and cosine functions multiplied by constants C₁ and C₂.

This universal form, y(t) = C₁ sin(kt) + C₂ cos(kt), shows how sine and cosine waves can be adjusted to satisfy the equation for any value of k. By choosing the appropriate values for C₁ and C₂, the general solution can be tailored to meet specific boundary conditions or initial values; these constants serve as the tailor's measurements to stitch the perfect fit.

Second-Order Linear Homogeneous Differential Equation

Peering into the world of differential equations, a second-order linear homogeneous differential equation has the standard form y''(t) + p(t)y'(t) + q(t)y(t) = 0, where p(t) and q(t) are functions of t. The 'homogeneous' part of this mouthful means that every term involves the function y or one of its derivatives; there's no stand-alone term just chilling by itself.

For the given problem, p(t) is missing (which means it's zero), and q(t) is a constant k². This simplicity allows us to use trigonometric functions as the underlying structure of the solutions. When k is any positive real number, the equation prefers the rhythms of sine and cosine waves to express its general solution. Like finding the exact notes in a musical composition, these waves hit the right frequency—multiply them by k—to resonate perfectly with the given differential equation.

Trigonometric Functions in Differential Equations

Why do trigonometric functions like sine and cosine show up in the conversation about differential equations? Imagine them as the essential building blocks that can model periodic phenomena, from the ebb and flow of ocean tides to the oscillations of a pendulum. In our specific equation, which has the form y''(t) + k^2y(t) = 0, the trigonometric functions sine and cosine are natural choices to form the solution because their derivatives loop back around to themselves, albeit with alternating signs or a scale factor.

This harmonic dance of signs and factors means that when you plug the derivatives into the equation, the terms cancel out perfectly, leaving zero—a clean, tidy solution. So, seeing y(t) = C₁ sin(kt) + C₂ cos(kt) should bring to mind a visualization of harmonious waves, perfectly in sync with the nature of the equation they're solving.

Derivative Calculation

Derivatives are all about change and calculating derivatives is like having a sixth sense for detecting how things are going to evolve over time. In the realm of differential equations, finding derivatives is crucial—it’s like gathering clues to solve a mystery. We take the original function—in this case, a combination of sine and cosine functions—and apply the rules of differentiation to find the first and second derivatives.

For our function y(t) = C₁ sin(kt) + C₂ cos(kt), the derivatives bring into play new versions of these original functions, but with a twist—each derivative switch hits you with a sign change or a k multiplier. By plugging these derivatives back into the equation, we complete the verification process, concluding that our sinusoidal suspects are indeed the culprits that satisfy the equation. This step of calculating and substituting derivatives serves as the final 'aha!' moment, confirming we've cracked the code.