Question

Finding a Particular Solution In Exercises \(15-18\) , find the function \(y=f(t)\) passing through the point \((0,10)\) with the given first derivative. Use a graphing utility to graph the solution. $$ \frac{d y}{d t}=\frac{3}{4} y $$

Short answer

The solution for the differential equation is \(y(t) = 10e^{\frac{3}{4}t}\).

Step-by-step solution

Set up the differential equation

Write down the given differential equation, which is \(\frac{d y}{d t} = \frac{3}{4} y\).

Separate and Integrate

Utilize the method of separating variables. Get all the \(y\)'s on one side and the \(t\)'s on the other. This results in \(\frac{d y}{y} = \frac{3}{4} dt\). Next, integrate both sides, resulting in \(\int \frac{d y}{y} = \int \frac{3}{4} dt\), which yields \(ln|y| = \frac{3}{4}t + C\).

Solve for \(y\)

To prepare for using the boundary condition, you need to solve for \(y\). To get \(y\) out of the natural logarithm, apply exponentiation, resulting in \(y = e^{\frac{3}{4}t + C}\).

Use the initial condition to solve for \(C\)

Apply the initial condition \((0,10)\), which means when \(t=0, y=10\). Substitute these values in to the equation from step 3 to get the constant \(C\): \(10 = e^{C}\), so \(C = ln(10)\).

Write out the final function

Now that \(C\) has been found, replace \(C\) in the equation from step 3 to get the final function: \(y(t) = e^{\frac{3}{4}t + ln(10)} = 10e^{\frac{3}{4}t}\).

Key concepts

Separation of Variables

When solving differential equations, separation of variables is a popular method used when the equation can be expressed in the form where all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side. This allows for easier integration of both sides.

In our example, we start with the differential equation \(\frac{d y}{d t} = \frac{3}{4} y\). To separate the variables, move all \(y\)-related terms to one side and \(t\)-related terms to the other. This results in \(\frac{d y}{y} = \frac{3}{4} dt\). By doing so, we have successfully separated the equation, which sets the stage for the next step: integration.

By integrating both sides, we aim to find the general solution. Remember that the left side involves \(y\) and the right side involves \(t\), which means we're separately dealing with distinct functions. Integrating \(\frac{d y}{y}\) gives us \(\ln|y|\), and integrating \(\frac{3}{4} dt\) gives us \(\frac{3}{4}t + C\), where \(C\) is the integration constant. This method is straightforward and powerful for this class of problems.

Initial Condition in Differential Equations

An initial condition is a crucial part of solving differential equations when looking for a particular solution. It provides a specific point that the broader solution set must satisfy. This information narrows down the infinite family of solutions to a unique solution.

In our scenario, the initial condition is given as \((0,10)\), indicating that when \(t = 0\), the function \(y\) takes the value of 10. After separating and integrating, we found that \(\ln|y| = \frac{3}{4}t + C\) leads to \(y = e^{\frac{3}{4}t + C}\).

Using the initial condition, substitute \(t = 0\) and \(y = 10\) into the equation, giving us \(10 = e^{C}\). Solving this gives us \(C = \ln(10)\). Plug this back into the general solution to get the particular solution. Hence, knowing the initial condition guides us to ward the exact solution for the problem.

Exponential Growth in Differential Equations

Exponential growth describes a process where the rate of change of a quantity is proportional to the current quantity itself. It's a common pattern found in differential equations. The solution often appears in the form \(y = Ce^{kt}\), where \(k\) is a constant.

In the given differential equation \(\frac{d y}{d t} = \frac{3}{4} y\), the rate of change of \(y\) with respect to \(t\) is directly proportional to \(y\) itself, with the proportionality constant \(\frac{3}{4}\). This implies exponential growth. After solving the equation, the solution \(y = 10e^{\frac{3}{4}t}\) clearly shows this growth behavior.

The constant \(10\) represents the initial amount when \(t = 0\), and the term \(e^{\frac{3}{4}t}\) captures the exponential growth over time. Such equations are widely used to model real-world scenarios like population growth or compound interest, where understanding the rate of increase is essential.