Solve The Initial-value Problem Dy/dt = 9y, Given Y(0) = 4. Find The Times When Y = 100 And Y = 1.

In this article, we will delve into the process of solving the initial-value problem (IVP) defined by the differential equation $\frac{dy}{dt} = 9y$, with the initial condition $y_0 = 4$. This type of problem is a cornerstone of differential equations, with applications spanning various fields, including physics, engineering, biology, and economics. Understanding how to solve these problems is crucial for modeling and predicting the behavior of dynamic systems. We will start by finding the general solution to the differential equation, then apply the initial condition to determine the specific solution. Finally, we will explore the times at which the solution reaches specific values, namely 100 and 1, rounding our answers to four decimal places.

Finding the General Solution

To begin, we need to find the general solution to the given differential equation, $\frac{dy}{dt} = 9y$. This is a first-order, linear, homogeneous differential equation, which can be solved using several methods, including separation of variables. The method of separation of variables is particularly effective for this type of equation. We rewrite the equation by separating the variables y and t:

$\frac{dy}{y} = 9 dt$

Now, we integrate both sides of the equation with respect to their respective variables:

$\int \frac{dy}{y} = \int 9 dt$

The integrals evaluate to:

$ln|y| = 9t + C$

where C is the constant of integration. To solve for y, we exponentiate both sides of the equation:

$e^{ln|y|} = e^{9t + C}$

$|y| = e^{9t}e^C$

Since $e^C$ is also a constant, we can replace it with another constant, let's call it A:

$y = Ae^{9t}$

This is the general solution to the differential equation. It represents a family of exponential functions, each determined by a specific value of the constant A. To find the particular solution that satisfies the given initial condition, we need to determine the value of A.

Applying the Initial Condition

The initial condition provided is $y_0 = 4$, which means that at time $t = 0$, the value of y is 4. We substitute these values into the general solution to solve for A:

$4 = Ae^{9(0)}$

$4 = Ae^0$

$4 = A(1)$

$A = 4$

Now that we have found the value of A, we can write the particular solution to the initial-value problem:

$y(t) = 4e^{9t}$

This is the unique solution that satisfies both the differential equation and the initial condition. It describes the exponential growth of y with respect to time t, starting from an initial value of 4. This specific solution allows us to predict the value of y at any given time t.

Determining the Time to Reach Specific Values

Time to Reach y = 100

Now, let's determine the time it takes for y to reach 100. We set $y(t) = 100$ and solve for t:

$100 = 4e^{9t}$

Divide both sides by 4:

$25 = e^{9t}$

Take the natural logarithm of both sides:

$ln(25) = ln(e^{9t})$

$ln(25) = 9t$

Solve for t:

$t = \frac{ln(25)}{9}$

Using a calculator, we find:

$t ≈ \frac{3.2189}{9} ≈ 0.3577$

Therefore, it takes approximately 0.3577 units of time for y to increase to 100.

Time to Drop to y = 1

Next, we determine the time it takes for y to drop to 1. We set $y(t) = 1$ and solve for t:

$1 = 4e^{9t}$

Divide both sides by 4:

$\frac{1}{4} = e^{9t}$

Take the natural logarithm of both sides:

$ln(\frac{1}{4}) = ln(e^{9t})$

$ln(\frac{1}{4}) = 9t$

Solve for t:

$t = \frac{ln(\frac{1}{4})}{9}$

Using a calculator, we find:

$t ≈ \frac{-1.3863}{9} ≈ -0.1540$

Therefore, it takes approximately -0.1540 units of time for y to drop to 1. The negative value indicates that y was equal to 1 at a time 0.1540 units before our initial time ($t = 0$). This result makes sense given the exponential growth nature of the solution; y was smaller in the past.

The solution to the initial-value problem $\frac{dy}{dt} = 9y$ with $y_0 = 4$ is:

$y(t) = 4e^{9t}$

The time it takes for y to increase to 100 is approximately:

$t ≈ 0.3577$

The time it takes for y to drop to 1 is approximately:

$t ≈ -0.1540$

In this article, we successfully solved the initial-value problem, found the specific solution, and determined the times at which the solution reaches specific values. This exercise demonstrates the power of differential equations in modeling dynamic systems and the importance of understanding the techniques for solving them. The exponential growth model we encountered here is a fundamental concept in many scientific and engineering applications. By mastering these concepts, you can analyze and predict the behavior of various real-world systems that exhibit exponential growth or decay.

• Initial-Value Problem (IVP): A differential equation coupled with an initial condition that specifies the value of the function at a particular point.
• Separation of Variables: A technique used to solve certain types of differential equations by isolating variables on different sides of the equation.
• General Solution: A family of solutions to a differential equation that includes an arbitrary constant.
• Particular Solution: A specific solution to a differential equation that satisfies a given initial condition.
• Exponential Growth: A phenomenon where a quantity increases at a rate proportional to its current value.
• Natural Logarithm: The logarithm to the base e, denoted as ln(x).

To deepen your understanding of differential equations and initial-value problems, consider exploring these topics further:

• Different types of differential equations: Learn about linear, nonlinear, homogeneous, and non-homogeneous equations.
• Numerical methods for solving differential equations: Explore techniques like Euler's method and Runge-Kutta methods.
• Applications of differential equations: Investigate how differential equations are used in physics, engineering, biology, economics, and other fields.

To solidify your understanding, try solving similar initial-value problems. For example:

1. Solve $\frac{dy}{dt} = -2y$ with $y(0) = 5$. At what time does y reach 0.5?
2. Solve $\frac{dy}{dt} = 4y$ with $y(0) = 2$. At what time does y reach 20?

By practicing these types of problems, you will develop a strong foundation in solving differential equations and modeling dynamic systems.

Solving initial-value problems is a fundamental skill in many areas of science and engineering. By understanding the concepts and techniques discussed in this article, you will be well-equipped to tackle a wide range of problems involving differential equations and dynamic systems. Remember to practice regularly and explore further topics to deepen your knowledge and expertise. The world of differential equations is vast and fascinating, and there is always more to learn!