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{d y}{d t}=4 y\)

Short answer

\( y = 100 \) at \( t = \frac{\ln 10}{4} \) and \( y = 1 \) at \( t = -\frac{\ln 10}{4} \).

Step-by-step solution

Understand the Problem

We need to solve the differential equation \(\frac{dy}{dt} = 4y\) with the initial condition \(y_0 = 10\) and determine the time \(t\) when \(y = 100\) or \(y = 1\).

Set Up the Differential Equation

The given differential equation is \(\frac{dy}{dt} = 4y\). This is a first-order linear differential equation that can be solved using the method of separation of variables.

Separate Variables and Integrate

To separate variables, we rewrite the equation as \(\frac{1}{y} dy = 4 dt\). Integrating both sides, we have \(\int \frac{1}{y} dy = \int 4 dt\).

Solve the Integrals

The integrals yield \(\ln |y| = 4t + C\), where \(C\) is the integration constant.

Solve for the Integration Constant

We apply the initial condition \(y_0 = 10\) (i.e., \(y(0) = 10\)) to find \(C\). Substituting into the integrated equation gives \(\ln |10| = 4(0) + C \Longrightarrow C = \ln 10\). Thus, the equation becomes \(\ln |y| = 4t + \ln 10\).

Solve for y(t)

Exponentiating both sides, we get \(|y| = e^{4t + \ln 10} = 10e^{4t}\). Since \(|y| = y\) in this context, the solution is \(y = 10e^{4t}\).

Find Time when y = 100

Set \(y = 100\) in the equation \(y = 10e^{4t}\), and solve for \(t\): \(100 = 10e^{4t} \Rightarrow e^{4t} = 10 \Rightarrow 4t = \ln 10 \Rightarrow t = \frac{\ln 10}{4}\).

Find Time when y = 1

Set \(y = 1\) in the equation \(y = 10e^{4t}\), and solve for \(t\): \(1 = 10e^{4t} \Rightarrow e^{4t} = \frac{1}{10} \Rightarrow 4t = -\ln 10 \Rightarrow t = -\frac{\ln 10}{4}\).

Key concepts

Initial Value Problems

Initial value problems, or IVPs, involve differential equations that require specific conditions to find a unique solution. In our exercise, we have the differential equation \(\frac{dy}{dt} = 4y\) with the initial condition \(y_0 = 10\). This means at the start, when time \(t = 0\), the function \(y\) has the value 10.
This initial condition allows us to determine a particular solution from the family of solutions of the differential equation. Without it, the solution would include an arbitrary constant, making it non-specific.

Separation of Variables

Separation of variables is a method used to solve first-order differential equations. This technique involves rearranging an equation so that all terms involving one variable are on one side, and all terms involving the other variable are on the opposite side.
In our example, the equation \(\frac{dy}{dt} = 4y\) is rearranged to \(\frac{1}{y} dy = 4 dt\). By doing this, we can integrate each side with respect to its own variable:

• Integrate \(\frac{1}{y} dy\) to get \(\ln|y|\).
• Integrate \(4 dt\) to get \(4t\).

This method simplifies the integration process and helps in finding the function \(y(t)\).

Exponential Growth

The term exponential growth describes a situation where the rate of change of a quantity is proportional to its current value. In our problem, \(\frac{dy}{dt} = 4y\), the rate at which \(y\) changes is a multiple of its current value \(4y\).
Such problems often involve solutions of the form \(y = Ce^{kt}\), where \(C\) is a constant and \(k\) is the growth rate. For our problem, after solving, we found \(y = 10 e^{4t}\). The constant \(10\) originated from the initial condition, showing how quickly \(y\) grows over time.

Integration Constant

An integration constant arises whenever you integrate a differential equation. It represents the family of functions that satisfy the indefinite integral. During solving, this constant is crucial when applying an initial condition to find a specific solution.
In our solution process, after integrating, we determined that \(\ln|y| = 4t + C\). To find \(C\), we used the initial condition \(y_0 = 10\) at \(t = 0\), yielding \(C = \ln 10\).

• This made our solution \(\ln|y| = 4t + \ln 10\).
• Finally, the exponential transformation led to \(y = 10e^{4t}\).

The integration constant helps tailor the general solution to specific conditions.