Problem 5 - Differential Equations and Linear Algebra

Question

Verify that, for \(t>0, y(t)=\ln t\) is a solution to the differential equation $$ 2\left(\frac{d y}{d t}\right)^{3}=\frac{d^{3} y}{d t^{3}} $$

Step-by-Step Solution

Answer

The first, second, and third derivatives of \(y(t) = \ln(t)\) with respect to t are \(\frac{dy}{dt} = \frac{1}{t}\), \(\frac{d^2 y}{dt^2} = \frac{-1}{t^2}\), and \(\frac{d^3 y}{dt^3} = \frac{2}{t^3}\), respectively. Plugging these into the given differential equation \(2\left(\frac{dy}{dt}\right)^3 = \frac{d^3 y}{dt^3}\), we get \(2\left(\frac{1}{t^3}\right) = \frac{2}{t^3}\), which confirms that \(y(t) = \ln(t)\) is indeed a solution to the given differential equation.

Step 1: Find the first derivative of y(t) with respect to t

To find the first derivative \( \frac{dy}{dt} \), we differentiate y(t) = ln(t) with respect to t: \[ \frac{dy}{dt} = \frac{1}{t} \]

Step 2: Find the second derivative of y(t) with respect to t

To find the second derivative \( \frac{d^2 y}{dt^2} \), we differentiate the first derivative with respect to t: \[ \frac{d^2 y}{dt^2} = \frac{-1}{t^2} \]

Step 3: Find the third derivative of y(t) with respect to t

To find the third derivative \( \frac{d^3 y}{dt^3} \), we differentiate the second derivative with respect to t: \[ \frac{d^3 y}{dt^3} = \frac{2}{t^3} \]

Step 4: Plug the first and third derivatives into the given differential equation

We will plug in the first and third derivatives of y(t) into the given differential equation: \[ 2\left(\frac{dy}{dt}\right)^3 = \frac{d^3 y}{dt^3} \] Substitute the derivatives we found in Steps 1 and 3: \[ 2\left(\frac{1}{t}\right)^3 = \frac{2}{t^3} \]

Step 5: Check if the equation is satisfied

Now, we check if the equation holds true after plugging in the derivatives: \[ 2\left(\frac{1}{t^3}\right) = \frac{2}{t^3} \] As we can see, both sides of the equation are equal. Therefore, y(t) = ln(t) is indeed a solution to the given differential equation.

Key Concepts

First Derivative

The journey to solving a differential equation typically begins with finding the first derivative, which tells us how a function is changing at any given point. Here, we're dealing with the function \( y(t) = \ln t \). When we differentiate this function with respect to \( t \), we use the basic rule of differentiation for logarithmic functions. The first derivative turns out to be:

• \( \frac{dy}{dt} = \frac{1}{t} \)

This result indicates that the rate of change of the logarithmic function \( \ln t \) is inversely proportional to \( t \). When \( t \) increases, the rate of change decreases, and vice versa. This first step sets the groundwork for finding subsequent derivatives.

Second Derivative

Once the first derivative is found, the next task is to explore the second derivative. This tells us about the "curvature" or the "concavity" of the original function \( y(t) \). In mathematical terms, it reflects how the rate of change itself is changing over time.To find the second derivative, \( \frac{d^2 y}{dt^2} \), we differentiate the first derivative \( \frac{1}{t} \) with respect to \( t \):

• \( \frac{d^2 y}{dt^2} = \frac{-1}{t^2} \)

The negative sign indicates that the function is concave downwards, suggesting a slowing rate of change. It shows that as \( t \) grows, the degree to which \( y(t) \) slows down increases.

Third Derivative

The third derivative provides further insight into how a function behaves. It describes the change in the curvature described by the second derivative. For our function, the sequence continues by differentiating the second derivative \( \frac{-1}{t^2} \) with respect to \( t \).This process gives us:

• \( \frac{d^3 y}{dt^3} = \frac{2}{t^3} \)

This result means that the way the curvature is changing is becoming more pronounced with increasing \( t \). The positive sign here shows increasing convexity in the curve, which is a vital component in verifying the differential equation.

Solution Verification

After computing the derivatives, the verification process is relatively straightforward. We need to check if the function \( y(t) = \ln t \), fulfills the given differential equation:

• \( 2\left(\frac{dy}{dt}\right)^3 = \frac{d^3 y}{dt^3} \)

To do this, substitute \( \frac{dy}{dt} = \frac{1}{t} \) and \( \frac{d^3 y}{dt^3} = \frac{2}{t^3} \) into the equation:

• \( 2\left(\frac{1}{t}\right)^3 = \frac{2}{t^3} \)

Both sides of the equation are equal, confirming that \( y(t) = \ln t \) is indeed a valid solution. This verification reassures us that the derived derivatives align perfectly with the requirements of the problem, ensuring there's no miscalculation.