Question

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$y^{\prime}(t)=\frac{t}{y}, y(1)=2$$

Short answer

Question: Given the initial value problem \(y'(t)=\frac{t}{y}\) with \(y(1)=2\), find the implicit solution to this problem. Answer: The implicit solution to the given initial value problem is \(\frac{1}{2}y^2 - \frac{1}{2}t^2 = \frac{3}{2}\).

Step-by-step solution

Separate Variables

First, let's rewrite the given differential equation as: $$y^{\prime}(t) y = t.$$ Now, we separate variables by moving all \(y\) terms to the left side and all \(t\) terms to the right side: $$y dy = t dt.$$

Integrate Both Sides

Now, we need to integrate both sides of the separated equation to solve for \(y(t)\): $$\int y dy = \int t dt.$$ On the left side, it is a simple integration concerning \(y\). On the right side, it is a simple integration concerning \(t\). The integration gives the following result: $$\frac{1}{2}y^2 = \frac{1}{2}t^2 + C,$$ where \(C\) is the constant of integration.

Find the Implicit Solution

To find the implicit solution, we can leave the equation like this: $$\frac{1}{2}y^2 - \frac{1}{2}t^2 = C.$$

Solve for the Constant of Integration

Now, let's use the initial value \(y(1)=2\) to find the constant \(C\). Plug in the initial value into the implicit solution: $$\frac{1}{2}(2)^2 - \frac{1}{2}(1)^2 = C.$$ Solving the equation gives us \(C=\frac{3}{2}\).

Final Implicit Solution and Graphing

With the constant of integration solved, we can now write down the final implicit solution for the initial value problem: $$\frac{1}{2}y^2 - \frac{1}{2}t^2 = \frac{3}{2}.$$ To graph this equation, we may use graphing software. The software will plot the curves representing the implicit solution, and we can identify the specific solution curve passing through the initial point \((1,2)\). Since the implicit solution represents a family of hyperbolas, be sure to choose the curve that passes through the given initial point.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They express relationships between changing quantities. In many real-world applications, these equations help us model phenomena such as population growth, heat transfer, and motion.

For the initial value problem presented, we start with a differential equation of the form \( y'(t) = \frac{t}{y} \). This equation relates the derivative of \( y \) concerning \( t \) to the variables \( t \) and \( y \) themselves.

The goal is to find a function \( y(t) \) that satisfies this relationship, given an initial condition like \( y(1)=2 \). This process often involves techniques like separation of variables and integration.

Separation of Variables

Separation of variables is a key technique for solving differential equations. It involves rearranging the equation so that each variable is on a different side. This allows us to solve each side separately through integration.

In this case, the differential equation \( y'(t) = \frac{t}{y} \) is rewritten by multiplying both sides by \( y \), giving \( y\,dy = t\,dt \).

This step is crucial because it transforms the problem into a pair of simpler equations, each depending on a single variable. At this point, each side can be integrated on its own, paving the way to the solution.

Implicit Solution

An implicit solution expresses a relationship between the variables without explicitly solving for one variable in terms of the other. This is sometimes preferred, especially when solving the equation explicitly is complex or impossible.

In the provided exercise, after separating variables and integrating, we arrived at the equation \( \frac{1}{2}y^2 - \frac{1}{2}t^2 = C \). This describes an implicit solution as \( y \) is not isolated.

The implicit form can represent multiple curves or solutions. For the initial value problem, we identify the specific curve using the initial condition, ensuring that the chosen solution satisfies it.

Integration

Integration is the process of finding a function whose derivative is the given function. It's an essential tool in calculus used to solve differential equations and find areas under curves.

Once the variables have been separated, each side of the equation \( y\,dy = t\,dt \) is integrated separately. The left side \( \int y\,dy \) results in \( \frac{1}{2}y^2 \), and the right side \( \int t\,dt \) gives \( \frac{1}{2}t^2 \).

• This process generates the implicit form \( \frac{1}{2}y^2 = \frac{1}{2}t^2 + C \).
• Constant terms are crucial here and appear as a result of the indefinite integration process.

Integration transforms the task from finding a specific relationship back to analyzing a broader, generalized one.

Constant of Integration

The constant of integration \( C \) is a fundamental part of solving indefinite integrals. Each indefinite integral yields a family of functions, differing by this constant.

In our differential equation, after integrating, we obtain \( \frac{1}{2}y^2 = \frac{1}{2}t^2 + C \).

This constant is determined using the initial condition \( y(1) = 2 \). Plugging \( t = 1 \) and \( y = 2 \) into the implicit solution yields \( C = \frac{3}{2} \).

• The initial value allows us to pinpoint the exact solution that satisfies the initial condition out of the infinite possibilities represented by the implicit solution.
• This critical step ensures the solution is not only general but also precisely tailored to the problem at hand.

Such understanding allows us to properly anticipate the behavior of the modeled system or situation.