Question

Use the ideas of Exercise 32 to solve the given initial value problem. Obtain an explicit solution if possible. $$ y^{\prime}=(t+y)^{2}-1, \quad y(1)=2 $$

Short answer

Answer: The explicit solution for the initial value problem is \(y(t) = \frac{te^{2t} - 2t + 2e^{2t} - 2}{e^{2t} - 1} - t\).

Step-by-step solution

Rewrite the ODE

We have, \(y'(t) = (t + y)^2 - 1.\) Rearrange the equation to show it in a separable form, i.e., \(f(y)dy = g(t)dt\). $$ \frac{dy}{dt} = (t + y)^2 - 1 \implies \frac{dy}{(t + y)^2 - 1} = dt $$

Separate variables and integrate both sides

We have separated variables, so now we integrate both sides of the equation. $$ \int\frac{dy}{(t + y)^2 - 1} = \int dt $$

Perform integration techniques

Let's perform a substitution to help with the integration. Let \(u = t + y\), then \(du = dy\). So, the equation becomes: $$ \int\frac{du}{u^2 - 1} = \int dt $$ Now, use partial fraction decomposition on the left side. We write \(\frac{1}{u^2-1} = \frac{A}{u-1} + \frac{B}{u+1}\), and find A and B: $$ A(u-1) + B(u+1) = 0 \implies A = -\frac{1}{2}, \; B = \frac{1}{2} $$ Integrating both sides, we have: $$ -\frac{1}{2}\int\frac{du}{u-1} + \frac{1}{2}\int\frac{du}{u+1} = \int dt $$

Calculate the integrals

Now, perform the integration: $$ \left[-\frac{1}{2}\ln\left|u-1\right| + \frac{1}{2}\ln\left|u+1\right|\right] = t + C $$

Substitute back for y and simplify

Recall that \(u = t + y\). Substitute back and apply properties of logarithms to simplify the equation: $$ \ln\left|\frac{t+y+1}{t+y-1}\right| = 2t + D $$ Note that when substituting back the bounds from step 3, the constant of integration might change, so it is denoted as D.

Apply the initial condition

Using the initial condition \(y(1) = 2\), apply the given values in the equation and solve for D: $$ \ln\left|\frac{1+2+1}{1+2-1}\right| = 2(1) + D \implies \ln\left|4\right| = 2 + D \implies D = \ln\left|4\right| - 2 $$

Solve for y(t)

Substitute the value of D into the equation and solve for y(t): $$ \ln\left|\frac{t+y+1}{t+y-1}\right| = 2t + \ln\left|4\right| - 2 $$ Equate inside the logarithm and isolate y(t): $$ \frac{t+y+1}{t+y-1} = e^{2t+ \ln\left|4\right| - 2} \implies y(t) = \frac{te^{2t} -2t + 2 e^{2t} - 2}{e^{2t} - 1} - t $$ We have now obtained an explicit solution for the initial value problem involving the given first-order ODE.

Key concepts

Differential Equations

Differential equations are mathematical models that describe the relationship between a function and its derivatives. They are central in expressing the dynamics of various systems, such as those in physics, engineering, and economics. A differential equation that involves a single independent variable is known as an ordinary differential equation (ODE). The general form of an ODE is \[ F(x, y, y', y'', ..., y^{(n)}) = 0, \]where \(y^{(n)}\) represents the \(n\)-th derivative of \(y\) with respect to \(x\).In the initial value problem given, we have a first-order ODE, indicated by the presence of only the first derivative of \(y\), \(y'\). The goal here is to find the function \(y(t)\) that satisfies the equation \(y'(t) = (t + y)^2 - 1\), subject to the condition \(y(1) = 2\). This initial condition ensures that we find a unique, specific solution to the equation.

Explicit Solution

An explicit solution to a differential equation is a formula that expresses the dependent variable as a function only of the independent variable without involving the derivative of the dependent variable. Essentially, an explicit solution takes the form \(y = f(t)\), allowing for direct computation of \(y\) for any given \(t\).In contrast, an implicit solution may involve both \(y(t)\) and \(t\) in an equation without isolating \(y(t)\) entirely. To obtain an explicit solution, the problem may require manipulation of the equation and the application of various mathematical techniques to separate and then integrate, such as was done in the given initial value problem. The final expression of \(y\) as a function of \(t\) in this problem is an instance of an explicit solution.

Integration Techniques

Integration techniques are a collection of methods used to evaluate integrals, which are essential to solving differential equations. Common integration techniques include substitution, integration by parts, partial fraction decomposition, trigonometric integration, and others. Developing proficiency in these techniques is vital in finding the solutions to many types of integrals that occur in differential equations.In the sample problem, a change of variables (\(u = t + y\)) simplifies the given integral. Partial fraction decomposition is then applied to separate the fraction into terms that are easier to integrate. Through these techniques, the integral becomes manageable, and its evaluation leads us closer to finding the explicit solution to the initial value problem.

Separable Differential Equations

Separable differential equations are a class of ordinary differential equations in which the terms involving the dependent variable \(y\) can be separated from the terms involving the independent variable \(t\). A separable differential equation can be expressed in the form \( f(y)dy = g(t)dt \), allowing for both sides of the equation to be integrated separately.To solve a separable ODE, one follows these general steps:

• Rearrange the equation to isolate \(dy\) on one side and \(dt\) on the other.
• Integrate both sides of the equation separately.
• Solve the resulting equation for \(y\) if possible to get an explicit solution.

In the exercise given, after rearranging terms and separating variables, the equation was successfully integrated on both sides. These steps allowed us to move from the general form of a separable differential equation to an explicit solution for the problem.