Question

Solve the following initial-value problems starting from $y_0=10.$ At what time does $y$ increase to 100 or drop to $1?\$\$\frac{dy}{dt}=e^{4t}$

Short answer

$y(t)$ will reach 100 at $t=\frac{1}{4}\ln (\frac{1361}{4})$, but it will never drop to 1.

Step-by-step solution

Identify the Differential Equation

We are given the differential equation $\frac{dy}{dt}=e^{4t}$, where $y_0=10$ is the initial value.

Integrate the Differential Equation

To find $y(t)$, integrate both sides of the equation $\frac{dy}{dt}=e^{4t}$. The integration of $e^{4t}$ with respect to $t$ is $\frac{1}{4}{e}^{4t}+C$. Thus, we have $y(t)=\frac{1}{4}{e}^{4t}+C$.

Apply the Initial Condition

Using the initial condition $y(0)=10$, substitute $t=0$ and $y=10$ into the integrated equation: $10=\frac{1}{4}{e}^0+C$. Solving for $C$, we find $C=10-\frac{1}{4}=\frac{39}{4}$.

Write the General Solution

Substitute $C=\frac{39}{4}$ back into the equation, resulting in $y(t)=\frac{1}{4}{e}^{4t}+\frac{39}{4}$.

Find Time When $y$ Equals 100

Set $y(t)=100$ and solve for $t$: $100=\frac{1}{4}{e}^{4t}+\frac{39}{4}$. Simplifying, $100-\frac{39}{4}=\frac{1}{4}{e}^{4t}$. Thus, $e^{4t}=400-\frac{39}{4}$. Solving gives $e^{4t}=\frac{1361}{4}$. Taking the natural logarithm, $4t=\ln (\frac{1361}{4})$. Divide by 4: $t=\frac{1}{4}\ln (\frac{1361}{4})$.

Find Time When $y$ Equals 1

Set $y(t)=1$ and solve for $t$: $1=\frac{1}{4}{e}^{4t}+\frac{39}{4}$. Simplifying, $1-\frac{39}{4}=\frac{1}{4}{e}^{4t}$. Thus, $e^{4t}=\frac{-35}{4}$. However, this value is not possible (since exponential functions are positive), so $y(t)$ will never drop to 1.

Key concepts

Differential Equations

Differential equations are essential tools in mathematics used to describe relations involving rates of change. They express these relationships using derivatives, which show how a given quantity changes with respect to another variable, often time. In this particular exercise, the differential equation given is $\frac{dy}{dt}=e^{4t}$. This expression highlights how the rate of change of $y$ with respect to time $t$ is linked to the exponential function $e^{4t}$.

Key ideas:

• The left side, $\frac{dy}{dt}$, represents the derivative, or rate of change, of $y$ over time.
• The right side, $e^{4t}$, dictates the specific dependency of this rate on time.
• Such equations are used in modeling dynamic systems, like population growth or radioactive decay, where changes over time are crucial.

This concept helps us understand how systems evolve over time when influenced by continuous compound effects represented by exponential functions.

Integration

Integration is a fundamental concept in calculus, essentially the reverse process of differentiation. It is used to determine a function from its derivative, revealing accumulative quantities. In this problem, integration helps us find $y(t)$, the function describing how $y$ changes over time.

Steps involved:

• Given $\frac{dy}{dt}=e^{4t}$, we integrate both sides with respect to $t$, resulting in $y(t)=\frac{1}{4}{e}^{4t}+C$.
• Here, $\frac{1}{4}{e}^{4t}$ arises because when integrating $e^{4t}$, we account for the coefficient of $t$ by multiplying by its reciprocal.
• The constant $C$ arises in indefinite integrals, representing the general solution to the differential equation.

This step is vital as it transitions from understanding how a system changes at each instant to describing the entire state of the system at any moment through $y(t)$.

Exponential Growth

Exponential growth describes a situation where the rate of growth of a quantity is proportionate to its current value, leading to rapid increases. The exercise illustrates this by showing how $y$ escalates exponentially over time. The differential equation $\frac{dy}{dt}=e^{4t}$ implies faster change as $t$ increases, typical of exponential functions.

Main characteristics:

• Exponential change means that the derivative, $\frac{dy}{dt}$, evolves as a power of $e$, leading to increasingly steep growth curves.
• For example, $y(0)=10$ sets a baseline, but as $t$ increases, $y(t)$ rapidly grows due to the influence of $e^{4t}$.
• Such growth models are common in natural phenomena, like population dynamics or financial interests, reflecting constant percentage increases over fixed periods.

This behavior underscores why the solution $y(t)=\frac{1}{4}{e}^{4t}+\frac{39}{4}$ grows without bound, never dropping to 1 as per the task, and highlights how exponential functions define intense acceleration in growth contexts.