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. $$ y^{\prime}+\frac{1}{y+1}=0, \quad y(1)=0 $$

Short answer

Question: Find the implicit solution to the given initial value problem, and check if the explicit solution exists. Also, determine the \(t\)-interval of existence if possible. Given initial value problem: $$ \frac{dy}{dt} = -\frac{1}{y+1}, \quad y(1)=0 $$ Answer: The implicit solution to the given initial value problem is $$ \frac{1}{2}y^2 + y + t = 1. $$ The explicit solution cannot be found in this case, and therefore the \(t\)-interval of existence cannot be determined.

Step-by-step solution

Write the given equation in differential form

Rewrite the given initial value problem as a separable first-order differential equation in differential form. $$ \frac{dy}{dt} = -\frac{1}{y+1} $$ Separate the variables, \(y\) and \(t\), to obtain the following $$ (y+1)dy = -dt. $$

Integrate both sides

Perform the integration on both sides of the equation: $$ \int (y+1) dy = - \int dt, $$ $$ \frac{1}{2}y^2 + y = - t + C. $$

Solve for the constant C

Apply the initial condition \(y(1)=0\) to solve for the integration constant \(C\): $$ \frac{1}{2}(0)^2 + 0 = - (1) + C. $$ So, \(C = 1\).

Write the implicit solution

Write down the implicit solution by inserting the value of \(C\) into the previous equation: $$ \frac{1}{2}y^2 + y + t = 1. $$

Check for the explicit solution (optional)

Attempt to isolate \(y\) on one side of the equation to determine the explicit solution. In this case, the equation $$\frac{1}{2}y^2 + y + t = 1$$ is not easily solvable for \(y\) explicitly. Hence, the explicit solution cannot be found.

Determine the t-interval of existence (optional)

Since we were unable to find an explicit solution for the initial value problem, it is not possible to determine the \(t\)-interval of existence.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in differential equations is a type of problem that not only involves finding a function that satisfies a given differential equation but also fulfills specific conditions at a starting point. It is crucial because it uniquely determines the solution by specifying an initial state of the system. This starting point is given as an initial condition. In our example, the initial condition given is \( y(1) = 0 \). This means that when \( t = 1 \), the value of \( y \) should be 0. This reduces the infinite possibilities of solutions to a unique one that exactly fits this requirement at \( t = 1 \).

Handling initial value problems effectively requires integrating both the differential equation and the initial condition, ensuring that the solution is tailored to the specific scenario described.

Implicit Solution

An Implicit Solution is a solution to a differential equation that is not presented in the form of \( y = f(t) \). Instead, the equation is written in a form that involves both \( y \) and \( t \). For example, an implicit solution might look like \( F(y, t) = 0 \).

In our problem, we derived the implicit solution \( \frac{1}{2}y^2 + y + t = 1 \). This equation involves both variables, and although it implicitly defines \( y \) in terms of \( t \), it cannot be easily rearranged to solve explicitly for \( y \). Implicit solutions are common when it's challenging or impossible to isolate a variable, and they still hold valuable solutions for many differential equations, capturing the relationship between the involved variables.

Separable Differential Equations

Separable Differential Equations are those that can be manipulated into a form where all terms involving one variable (and its differential) are on one side of the equation, and all terms involving the other variable (and its differential) are on the other. This split allows for straightforward integration. In our example, we transformed the equation \( \frac{dy}{dt} = -\frac{1}{y+1} \) by separating variables into \((y+1) dy = -dt\).

By achieving this separation, we can integrate both sides independently—one as a function of \(y\) and the other as a function of \(t\). This technique simplifies solving complex equations by reducing them into integrable parts, showcasing its power in tackling initial value problems efficiently.

Integration Techniques

Integration Techniques are essential tools in solving differential equations, especially when dealing with separable ones. Once the variables \( y \) and \( t \) have been separated, as seen in the transformation to \((y+1) dy = - dt\), integration allows us to solve each side as an independent integrable part.

For instance, the left side, \( \int (y+1) dy \), uses the basic rule of integrating linear polynomials and yields \( \frac{1}{2}y^2 + y \). Simultaneously, integrating \( -\int dt \) results in \( -t \). The solution is then combined to form an implicit solution.

• Techniques like substitution, partial fractions, or integration by parts could be employed if the functions were more complex.
• In practical applications, integrating constants is essential as they adjust the function to match the given initial conditions.

Understanding these techniques helps students solve a wide range of problems using calculus, enabling the transformation of differential equations into usable solutions.