Question

In each exercise, (a) Show that the given differential equation is not exact. (b) Multiply the equation by the function \(\mu\), if it is provided, and show that the resulting differential equation is exact. If the function \(\mu\) is not given, use the ideas of Exercise 22 to determine \(\mu\). (c) Solve the given problem, obtaining an explicit solution if possible. \((2 t+y) y^{\prime}+y=0, \quad y(2)=-3\)

Short answer

How would you determine the integrating factor and solve for the function y(t)? Answer: The given differential equation is not exact. To find the integrating factor, we need to compute the ratio \(\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}\right)\). In this case, the integrating factor μ depends on both t and y, so we cannot find the explicit solution for the function y(t).

Step-by-step solution

Verify if the given differential equation is exact

Write the given differential equation as \(M + N y^\prime = 0\), where \(M(t, y) = y\) and \(N(t,y) = 2t + y\). Check if the equation is exact by computing \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\): $$\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y) = 1$$ $$\frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(2t + y) = 2$$ Since \(\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}\), the given differential equation is not exact.

Determine the integrating factor μ

To find the integrating factor μ, let's compute the ratio \(\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}\right)\): $$\frac{1}{N}\left(\frac{\partial M}{\partial y} - \frac{\partial N}{\partial t}\right) = \frac{1}{2t + y}(1 - 2) = -\frac{1}{2t+y}$$ This suggests that the integrating factor μ is a function of t only: $$\mu(t) = \exp\left(\ -\int\frac{dt}{2t+y}\right)$$ Since the expression involves y, we can't find μ explicitly. However, given its general form, we can proceed to solve the problem using the proposed integrating factor.

Multiply the equation by μ and find the exact solution

Multiply the given differential equation by μ: $$(2t + y) y'\mu + y\mu = 0$$ Such that \(\mu\) is an integrating factor. Now the equation is exact: $$\frac{\partial}{\partial y}(y\mu) = \mu \Rightarrow \frac{\partial}{\partial t}[(2t + y)\mu] = \mu$$ Now, integrate both sides with respect to t: $$\int(2t + y)\mu\,dt = \int \mu\,dt$$ Since we could not find the explicit expression for μ, we cannot proceed with integration. However, we can conclude that integrating both sides would give us the implicit solution for y(t).

Identify the general solution and apply the initial condition

Assuming we could complete step 3, we would have found the general solution, say Y(t). To find the constant of integration, we apply the initial condition y(2) = -3: $$Y(2) = -3$$ Solving for the constant of integration, we would have obtained the explicit solution for the function y(t). However, due to the integrating factor's dependence on y, we could not find the explicit integrating factor or the explicit solution in this particular case.

Key concepts

Exact Differential Equation

Understanding what an exact differential equation is can be quite elemental in solving many mathematical problems. In essence, it's a type of differential equation that can be expressed in the form \(M(t, y) + N(t, y)y' = 0\) where \(M\) and \(N\) are functions of \(t\) and \(y\). The crux of these equations is that there exists a function \(F\) such that \(F_t = M\) and \(F_y = N\), meaning the partial derivatives of \(F\) with respect to \(t\) and \(y\) are \(M\) and \(N\) respectively. If a differential equation meets this criterion, we can integrate to find \(F\), which leads us to the solution of the equation. However, when \(\frac{\partial M}{\partial y} eq \frac{\partial N}{\partial t}\), the equation is not exact, and other methods, like finding an integrating factor, must be used. This caveat leads us to our next concept.

Integrating Factor

An integrating factor is a lifesaver when dealing with non-exact differential equations. It's a function, typically denoted as \(\mu\), that when multiplied by the original differential equation, transforms it into an exact equation. The process involves a bit of detective work since we need to find a \(\mu\) that meets the requirements. The clever part about an integrating factor is that it doesn't change the solution of the differential equation; it merely puts it in a form that we can solve using standard techniques. Normally, we'd look for \(\mu\) as a function of \(t\) or \(y\) only, and it's derived from a particular condition based on \(M\) and \(N\). In our textbook example, though, we stumbled upon a snag since \(\mu\) appeared to be a function of both \(t\) and \(y\), meaning we couldn't find it explicitly. This predicament nudges us towards an implicit solution, another intriguing topic.

Implicit Solution

When an equation intertwines variables so closely that they cannot be neatly separated, we end up with an implicit solution. This type of solution is expressed in a form where the dependent variable \(y\) is not isolated on one side of the equation; both \(y\) and the independent variable \(t\) are mixed on the same side. The implicit solution is still legitimate and often pops up when solving exact differential equations or when an integrating factor comes into play, and we can’t disentangle \(y\) fully. If we had been able to integrate after applying the integrating factor in our exercise, the result would be an expression that relates \(t\) and \(y\) without explicitly solving for \(y\), hence the term 'implicit solution'.

Explicit Solution

Contrary to an implicit solution, an explicit solution is the gold standard that most students aim for as it allows us to express the dependent variable \(y\) solely as a function of the independent variable \(t\): \(\frac{\text{dy}}{\text{dt}} = f(t)\). In the ideal world, every differential equation would come with an explicit solution, neatly separating \(y\) on one side to elegantly reveal the function's behaviour over time. Unfortunately, in the wild world of differential equations, obtaining such clear-cut solutions is not always possible. Sometimes we're bound to accept implicit solutions, though for many practical purposes, being able to represent \(y\) explicitly offers a simplified and more straightforward understanding of the relationship between variables.

Initial Value Problem

Much like how a seed grows into a plant given the right environment, in mathematics, an initial value gives birth to a unique solution in a differential equation's domain. An initial value problem (IVP) consists of a differential equation alongside an initial condition which specifies the value of the function at a particular point. In our textbook exercise, the initial condition given was \(y(2) = -3\), providing the necessary starting point to determine the constant of integration, which is essential once we have found a general solution. Solving an IVP allows us to 'personalize' the general solution to a particular scenario, offering a precise description of the system's behaviour from that initial point onwards. Since we couldn't find an explicit integrating factor or solution in the exercise, our capacity to solve the IVP was also limited. However, in typical cases, applying the initial condition is what brings us from a general idea to a precise, clear-cut answer to a problem.