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. $$ y^{\prime}-2 t y^{2}=0, \quad y(0)=-1 $$

Short answer

Answer: The explicit solution for the given IVP is \(y(t) = -\frac{1}{t^2 + 1}\).

Step-by-step solution

Separating Variables

First, we need to rewrite the given differential equation in a separable form, so that we can integrate both sides with respect to \(t\): $$ y^{\prime}-2 t y^{2}=0 $$ We can rewrite this equation as: $$ \frac{y^\prime}{y^2} = 2t $$

Integration

Now we can integrate both sides with respect to \(t\): $$ \int \frac{1}{y^2} \, dy^{\prime} = \int 2t \,dt $$ From this, we obtain: $$ -\frac{1}{y} = t^2 + C $$

Solving for the Implicit Solution

The implicit solution is: $$ -\frac{1}{y} = t^2 + C $$

Applying the Initial Condition

Now we apply the initial condition \(y(0)=-1\): $$ -\frac{1}{-1} = 0^2 + C $$ $$ C = 1 $$ So the implicit solution is: $$ -\frac{1}{y} = t^2 + 1 $$

Finding the Explicit Solution

Solving for \(y\), we get the explicit solution: $$ y(t) = -\frac{1}{t^2 + 1} $$

Determining the t-interval of existence

As \(y(t) = -\frac{1}{t^2 + 1}\), there are no singularities or points at which the function is not defined in the domain of \(t \in (-\infty, \infty)\). Therefore, the t-interval of existence is: $$ t \in (-\infty, \infty) $$

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are crucial in modeling situations where change is continuous, such as in physics, engineering, and economics. The equation provided in the exercise, \( y' - 2ty^2 = 0 \), is an example of a first-order differential equation because it involves the first derivative of the function \( y \) with respect to the variable \( t \) and can describe the rate of change of a variable.

An initial value problem in differential equations is where you are given a differential equation along with conditions that specify the value of the function and its derivatives at a certain point. This information is used to find a specific solution to the differential equation that not only satisfies the equation itself but also the initial conditions provided.

Separable Variables

A differential equation is said to have separable variables if it can be written in such a way that all terms involving one variable are on one side of the equation and all terms involving the other variable are on the opposite side. The equation \( y' - 2ty^2 = 0 \) can be manipulated into the form \( \frac{y'}{y^2} = 2t \), which allows us to integrate both sides separately.

Integration of Separable Variables

Once we have separated the variables, we integrate each side with respect to its own variable to find a solution. In our case, integrating \( \frac{1}{y^2} dy' \) gives us \( -\frac{1}{y} \) on the left, and integrating \( 2t dt \) gives us \( t^2+C \) on the right, where \( C \) is an integration constant.

Implicit Solution

An implicit solution to a differential equation represents the solution in a form where the dependent variable (typically \( y \)) is not isolated on one side of the equation. After integrating a separable equation, we typically get an implicit solution. From our exercise, after performing the integration as shown in the solution, the implicit solution is \( -\frac{1}{y(t)} = t^2 + C \).

This equation involves both \( y \) and \( t \) and does not explicitly solve for \( y \) as a function of \( t \). Finding an implicit solution is a useful step in the process of solving differential equations, as it can often lead to finding an explicit solution.

Explicit Solution

Unlike implicit solutions, an explicit solution is one where the dependent variable is expressed only in terms of the independent variable, with no other terms on the same side of the equation. For our problem, the explicit solution is acquired by manipulating the implicit form \( -\frac{1}{y(t)} = t^2 + 1 \) to solve for \( y(t) \) directly, yielding \( y(t) = -\frac{1}{t^2 + 1} \).

Benefits of Explicit Solutions

Explicit solutions are beneficial as they directly relate the dependent variable to the independent variable, making it easy to graph and analyze their behavior. In this form, predicting the system's progression based on different values of \( t \) becomes straightforward.

T-Interval of Existence

The t-interval of existence refers to the range of values for the independent variable \( t \) over which the solution to a differential equation is defined. Determining this interval is crucial as it tells us where the solution is valid and applicable. For the explicit solution \( y(t) = -\frac{1}{t^2 + 1} \), since the denominator never equals zero for any real number \( t \) and the function is defined for all real \( t \) values, the t-interval of existence is \( t \) in \( (-\infty, \infty) \).

The absence of singularities or discontinuities implies that the behavior of the solution can be reliably predicted for all time intervals, which is often desired in physical and empirical models.