Question

(a) Obtain an implicit solution and, if possible, an explicit solution of the initial value problem. (b) If you can find an explicit solution of the problem, determine the \(t\)-interval of existence. $$ \frac{d y}{d t}=\frac{1}{y^{2}}, \quad y(1)=2 $$

Short answer

Question: Solve the initial value problem given by the first-order ordinary differential equation \(\frac{d y}{d t} = \frac{1}{y^2}\) with an initial condition \(y(1) = 2\). Find the implicit and explicit solutions and the t-interval of existence. Answer: The implicit solution to the given ODE is \(-\frac{1}{y} = \frac{1}{y^{2}}(t - C_1)\). The explicit solution is \(y(t) = \frac{-1}{t - \frac{1}{2}}\). The t-interval of existence is \(t \in (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\).

Step-by-step solution

(Step 1: Solve the ODE for the implicit solution)

To find the implicit solution, we need to integrate both sides of the ODE with respect to \(t\): $$ \int \frac{d y}{d t} dt = \int \frac{1}{y^{2}} dt $$ Using the substitution \(u = y\), we obtain: $$ \int du = \int \frac{1}{u^2} dt $$ Integrating both sides, we get: $$ -\frac{1}{u} = \frac{1}{u^{2}}\int dt + C_1 $$ Replacing \(u\) with \(y\), we obtain the implicit solution: $$ -\frac{1}{y} = \frac{1}{y^{2}}(t - C_1) $$

(Step 2: Solve the ODE for the explicit solution)

To find the explicit solution, we will try to isolate \(y\) in the expression we found in step 1. $$ y^2(-\frac{1}{y}) = y^2(t - C_1) $$ $$ -y = y^2(t - C_1) $$ $$ y + y^2(t - C_1) = 0 $$ We have a quadratic equation in terms of \(y\). However, solving this equation for \(y\) may not yield an explicit solution. Considering that \(y = \frac{-1}{t - C_1}\) has a similar structure and might be a correct guess, let's plug it in into our implicit equation and see if it holds: $$ -\frac{1}{\frac{-1}{t - C_1}} = \frac{1}{(\frac{-1}{t - C_1})^2}(t - C_1) $$ $$ t - C_1 = t - C_1 $$ Our guess for the explicit solution of \(y = \frac{-1}{t - C_1}\) verifies the implicit equation, thus it is an explicit solution.

(Step 3: Apply the initial condition)

Now, let's use the initial condition \(y(1)=2\) to find the constant \(C_1\): $$ 2 = \frac{-1}{1 - C_1} $$ $$ C_1 = \frac{-1}{2} + 1 $$ $$ C_1 = \frac{1}{2} $$ Thus, our explicit solution is: $$ y(t) = \frac{-1}{t - \frac{1}{2}} $$

(Step 4: Determine the t-interval of existence)

Our explicit solution is defined for all \(t\) except for when the denominator is equal to 0: $$ t - \frac{1}{2} \neq 0 $$ So, the \(t\)-interval of existence is: $$ t \in (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty) $$

Key concepts

Implicit Solution

Understanding an implicit solution is a key step in solving differential equations. An implicit solution to a differential equation includes the function and its derivative interwoven together in an equation and not isolated. For the student grappling with the formation of an implicit solution, it's helpful to remember that the solution describes a relationship between the variables without explicitly solving for the dependent variable.

In our initial value problem, \(\frac{d y}{d t}=\frac{1}{y^{2}}\), the implicit solution is reached by integrating both sides. Integration, in this context, is the process of finding a function that describes the accumulation of quantities, such as areas under a curve. After integration and using substitution, we derived the implicit form \(-\frac{1}{y} = \frac{1}{y^{2}}(t - C_1)\). This equation shows the relationship between \(y\) and \(t\) without giving a direct formula for \(y\) as a function of \(t\). An implicit solution might not always lead to a direct extraction of the dependent variable, but it is a crucial step towards fully understanding the behavior of the differential equation.

Explicit Solution

Contrasting with the implicit solution is the explicit solution, where the dependent variable is isolated on one side of the equation. It directly expresses the dependent variable as a function of the independent variable, which in this case is \(y(t)\).

Crafting an explicit solution may involve manipulation of the implicit form, aiming to represent \(y\) solely in terms of \(t\). However, it’s not always possible to find such a solution. For the initial value problem given, an explicit solution was obtained by making an educated guess based on the structure of the implicit solution and then testing that guess. Eventually, we arrived at the explicit form \(y(t) = \frac{-1}{t - \frac{1}{2}}\) by satisfying the implicit equation and the given initial condition \(y(1)=2\). This explicit expression is useful for students in understanding the behavior of \(y\) throughout the range of independent variable \(t\), providing an immediate calculation of \(y\) for any value of \(t\) in its domain.

Differential Equations

Differential equations form the backbone of dynamic systems analysis. They are a form of mathematical equations that relate some function with its derivatives. They signify the rate at which changes occur, making them essential in fields like physics, engineering, economics, and biology.

The initial value problem provides us with a differential equation \(\frac{d y}{d t}=\frac{1}{y^{2}}\) alongside a starting condition. To interpret this equation, imagine \(y\) as a position over time \(t\), and \(\frac{d y}{d t}\) as the velocity. The equation thus describes the velocity in terms of the position squared. It reveals how the position changes in time and how the nature of this change depends on the position itself. In teaching, it is vital to emphasize the real-world phenomena these equations can model, such as the growth of populations or how electric circuits behave over time. Providing tangible examples helps students to anchor these abstract concepts in reality.

t-interval of Existence

The \(t\)-interval of existence is a concept that denotes the range of the independent variable—usually time—for which the solution to a differential equation is valid. It is essential for students to understand that not all functions are valid everywhere; some have limits or boundaries of applicability.

When dealing with the explicit solution \(y(t) = \frac{-1}{t - \frac{1}{2}}\), the function is undefined at \(t = \frac{1}{2}\) because it would result in division by zero, which is undefined mathematical operation. Therefore, the \(t\)-interval of existence excludes this point, and the function is said to exist for all \(t \) except \(\frac{1}{2}\). This concept is portrayed by the expression \(t \in (-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\). Instructing students about the importance of the \(t\)-interval of existence not only helps in understanding the conditions under which solutions apply but also aids in the deeper grasp of how solutions behave in the context of the problem at hand.