Question

Verifying solutions of initial value problems Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(t)=-3 \cos 3 t ; y^{\prime \prime}(t)+9 y(t)=0, y(0)=-3, y^{\prime}(0)=0$$

Short answer

In conclusion, the given function \(y(t) = -3 \cos 3t\) is a solution of the initial value problem with the differential equation \(y''(t) + 9y(t) = 0\) and initial conditions \(y(0) = -3\) and \(y'(0) = 0\). We have verified this by computing the first and second derivatives of the given function, checking that they satisfy the differential equation and the initial conditions.

Step-by-step solution

Compute the first derivative of the given function

To find the first derivative of the given function, we apply the chain rule, which states that for any function \(y(t) = f(g(t))\), the first derivative \(y'(t) = f'(g(t))\cdot g'(t)\). Let \(f(u) = -3\cos u\) and \(g(t) = 3t\). Then \(y(t) = f(g(t))\). The derivatives of \(f\) and \(g\) are: $$f'(u) = \frac{d}{du}(-3\cos u) = 3\sin u$$ $$g'(t) = \frac{d}{dt}(3t) = 3$$ Applying the chain rule, we find the first derivative of the given function: $$y'(t) = f'(g(t))\cdot g'(t) = 3\sin(3t) \cdot 3 = 9\sin(3t)$$

Compute the second derivative of the given function

Now, we need to find the second derivative of the given function, \(y''(t)\). We find the second derivative by taking the derivative of the first derivative. The first derivative is \(y'(t) = 9\sin(3t)\). Let \(h(v) = 9\sin v\), with \(f(t) = 3t\). Then \(y'(t) = h(f(t))\). Take the derivatives of \(h\) and \(f\): $$h'(v) = \frac{d}{dv}(9\sin v) = -9\cos v$$ $$f'(t) = \frac{d}{dt}(3t) = 3$$ Applying the chain rule, we find the second derivative: $$y''(t) = h'(f(t))\cdot f'(t) = -9\cos(3t) \cdot 3 = -27\cos(3t)$$

Check for the differential equation

Now that we have both the first and second derivatives, we can check if the given function satisfies the given differential equation \(y''(t) + 9y(t) = 0\). Plugging the expressions of \(y(t)\) and \(y''(t)\) into the differential equation, we have: $$-27\cos(3t) + 9(-3\cos(3t)) = 0$$ $$-27\cos(3t) - 27\cos(3t) = 0$$ $$-54\cos(3t) = 0$$ The equation holds true for all values of \(t\) so the function \(y(t)\) satisfies the given differential equation.

Check the initial conditions

Now, we have to check if the given function and its derivatives satisfy the initial conditions \(y(0) = -3\) and \(y'(0) = 0\). For the first initial condition, we plug \(t = 0\) into the given function: $$y(0) = -3\cos(3 \cdot 0) = -3\cos(0) = -3(1) = -3$$ The first initial condition is satisfied. For the second initial condition, we plug \(t = 0\) into the first derivative: $$y'(0) = 9\sin(3 \cdot 0) = 9\sin(0) = 9(0) = 0$$ The second initial condition is satisfied as well.

Conclusion

Since the given function and its derivatives satisfy both the differential equation and the initial conditions, we can conclude that \(y(t) = -3 \cos 3t\) is indeed a solution of the given initial value problem.

Key concepts

Differential Equation

A differential equation involves derivatives of a function. In simpler terms, it's an equation that relates a function with its derivatives. In our exercise, the differential equation is given as \( y''(t) + 9y(t) = 0 \). Here, \( y''(t) \) is the second derivative of the function \( y(t) \).
The purpose of solving a differential equation is to find a function \( y(t) \) that satisfies this equation for all values of \( t \).
In this problem, we are given a candidate function \( y(t) = -3 \cos 3t \) and we need to check if it satisfies the differential equation. When we plug in the function and its second derivative into the equation, it must hold true for all \( t \).
Solving differential equations involves understanding derivatives and knowing how to manipulate them to verify solutions.

Chain Rule

The chain rule is a fundamental rule in calculus used to differentiate composite functions. It helps us find the derivative of a function composed of another function.
In our exercise, we apply the chain rule to find the first derivative of the given function \( y(t) = -3\cos 3t \).
• To solve this, let \( f(u) = -3\cos u \) and \( g(t) = 3t \). This makes \( y(t) \) a composite function \( f(g(t)) \).
• The chain rule tells us that the derivative \( y'(t) \) is \( f'(g(t))\cdot g'(t) \).
• So, differentiating each part gives \( f'(u) = 3\sin u \) and \( g'(t) = 3 \), leading to \( y'(t) = 9\sin(3t) \).
The same process is applied again to find the second derivative.

Initial Conditions

Initial conditions are specific values of a function and its derivatives at a particular point, often used to determine a unique solution to a differential equation. They tell us what happens at the outset or initial moment of the scenario described.
In our problem, there are two initial conditions: \( y(0) = -3 \) and \( y'(0) = 0 \).
• For \( y(0) = -3 \): Substitute \( t = 0 \) into \( y(t) = -3\cos 3t \) which becomes \( -3 \cos 0 = -3 \).
• For \( y'(0) = 0 \): Substitute \( t = 0 \) into \( y'(t) = 9\sin 3t \) which becomes \( 9\sin 0 = 0 \).
Each satisfies the respective initial condition, confirming the particular behavior of the function and supporting its candidacy as a solution.

Second Derivative

The second derivative, expressed as \( y''(t) \), gives us information about the curvature or concavity of the function graph. It is simply the derivative of the derivative, providing insights into the change of the rate of change of the function.
In the context of this problem, we need the second derivative to check whether the function satisfies the differential equation \( y''(t) + 9y(t) = 0 \).
• From the first derivative \( y'(t) = 9\sin 3t \), we find \( y''(t) = -27\cos 3t \) using the chain rule again.
• Substituting into the equation gives \( -27\cos 3t + 9(-3\cos 3t) \), which simplifies to zero, proving the function \( y(t) \) meets the differential equation.
Thus, the second derivative is instrumental in verifying the candidate function.