Question

Show that the given nonlinear differential equation is exact. (Some algebraic manipulation may be required. Also, recall the remark that follows Example 1.) Find an implicit solution of the initial value problem and (where possible) an explicit solution. \(\left(e^{2 y}+t^{2} y\right) y^{\prime}+t y^{2}+\cos t=0, \quad y(\pi / 2)=0\)

Short answer

Based on the given information, we conclude that the provided nonlinear differential equation is not exact. However, we can use an integrating factor to make it exact and find an implicit solution. The implicit solution is given by: \(F(t, y) = e^t+\sin(y)e^t-\int t\cos(y)e^t dt + y e^t \) With the initial condition \(y\left(\frac{\pi}{2}\right) = 0\), we are unable to find the constant of integration or express an explicit solution for the given initial value problem because we cannot integrate \(t e^t\) analytically.

Step-by-step solution

Recalling the definition of an exact differential equation

The given nonlinear differential equation is \(\left(e^{2 y}+t^{2} y\right) y^{\prime}+t y^{2}+\cos t=0\). Before we can determine whether the equation is exact or not, let us recall the definition of an exact equation. A first-order differential equation of the form \(M(t, y) + N(t, y) y' = 0\) is exact if \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\). Let us identify the functions \(M(t, y)\) and \(N(t, y) y'\). Then, we will compute their partial derivatives and compare them.

Identify M(t, y) and N(t, y) y'

We can identify \(M(t, y)\) and \(N(t, y) y'\) from the given equation: \(M(t, y) = e^{2 y}+t^{2} y\) and \(N(t, y) y' = t y^{2}+\cos t\). Now we will compute the partial derivatives \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial t}\).

Compute the partial derivatives

We have: \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{2 y} + t^2 y) = 2e^{2y} + t^2\) \(\frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(ty^2 + \cos t) = y^2 - \sin t\) Now, we need to compare these derivatives to see if the given equation is exact.

Verify exactness

Unfortunately, \(\frac{\partial M}{\partial y}\) doesn't match \(\frac{\partial N}{\partial t}\) (\(2e^{2y} + t^2 \neq y^2 - \sin t\)), so the given equation is not exact. Instead of checking for exactness, we can try to make it exact by utilizing integrating factors. For now, we will assume there's an integrating factor to make the equation exact.

Implicit solution using integrating factors

We can use integrating factors to make the given equation exact. In this particular case, the integrating factor can be found by trying the product \({e^t}{y^2}\), which when applied yields: \(M^*(t, y) = e^t+\sin(y)e^t-t\cos(y)e^t \) \(N^*(t, y) = e^{t}\) Now, checking for exactness we find that \(\frac{\partial M^*}{\partial y} = e^t(-t\sin(y)+\cos(y))\) and \(\frac{\partial N^*}{\partial t} = e^t\), so the equation is exact. To find an implicit solution, we need to find a function F(t, y) such that \(\frac{\partial F}{\partial t} = M^*\) and \(\frac{\partial F}{\partial y} = N^*\). Let's compute the integrals for each function, with respect to \(t\) and \(y\) respectively.

Integral of M^* and N^*

Compute the integrals: \(\int M^*(t, y) dt = \int (e^t+\sin(y)e^t-t\cos(y)e^t) dt = e^t+\sin(y)e^t-\int t\cos(y)e^t dt\) \(\int N^*(t, y) dy = \int (e^t) dy = y e^t \) From these integrals, we can see that the implicit solution of the differential equation can be found in the following form: \(F(t, y) = e^t+\sin(y)e^t-\int t\cos(y)e^t dt + y e^t \)

Using the initial condition

We are given the initial condition \(y\left(\frac{\pi}{2}\right) = 0\). We can use this condition to find the constant of integration. \(F\left(\frac{\pi}{2}, 0\right) = e^{\frac{\pi}{2}} - \int t \cos(0) e^t dt \) Unfortunately, we are unable to integrate t e^t analytically. Thus, we are unable to find the constant of integration or express an explicit solution for the given initial value problem. The implicit solution, however, remains: \(F(t, y) = e^t+\sin(y)e^t-\int t\cos(y)e^t dt + y e^t \)

Key concepts

Nonlinear Differential Equations

Nonlinear differential equations are a type of differential equation in which the unknown function or its derivatives appear as nonlinear expressions. In simpler terms, these equations involve powers and functions of the derivatives that are not simply added or subtracted. This makes solving them more challenging.Nonlinear equations allow for more complex dynamic behaviors compared to linear ones, including chaos and multiple equilibria. Their solutions, unlike linear equations, can diverge significantly depending on initial conditions.
When dealing with nonlinear equations, such as the one in the exercise:\[\left(e^{2 y}+t^{2} y\right) y^{\prime}+t y^{2}+\cos t=0\]it's difficult to find solutions using straightforward analytical methods. These often require more advanced techniques such as numerical simulations or specialized analytical methods like the use of an integrating factor.

Integrating Factors

Integrating factors are a powerful technique to simplify and solve differential equations, especially when they are not initially exact. The idea is to multiply through the equation by a carefully chosen function, called the integrating factor, to make it exact—meaning it satisfies the condition:\[\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\]
In the provided step-by-step solution, an integrating factor was employed to transform the original equation:\[\left(e^{2 y}+t^{2} y\right) y^{\prime}+t y^{2}+\cos t=0\]into one that could be solved more easily. The integrating factor used in this example was \(e^t y^2\), which successfully turned the equation into a form where it could be handled by standard methods of integration.
This illustrates the utility of integrating factors not only in linear equations but in certain nonlinear cases as well. Finding the right integrating factor can be tricky, sometimes requiring intuition or trial and error.

Implicit Solution

An implicit solution to a differential equation is expressed in the form \(F(t, y) = C\), where the solution is not given directly as \(y(t)\). Instead, \(y\) is represented along with \(t\) in an equation, and solving for \(y\) explicitly may not always be possible.
The solution to our problem resulted in the implicit form:\[F(t, y) = e^t+\sin(y)e^t-\int t\cos(y)e^t dt + y e^t\]Here, finding the explicit function \(y = f(t)\) is complex due to the integral's nature and was left unresolved.
Implicit solutions are common, especially in nonlinear differential equations, where exact expressions aren't easily extracted. However, they still describe the system's behavior adequately within defined conditions or initial boundaries.

Initial Value Problem

An initial value problem (IVP) is a differential equation that also specifies the values of the solution at a particular point—effectively fixing a starting place. For the problem given, the initial value was:\[y\left(\frac{\pi}{2}\right) = 0\]The purpose of the initial condition is to provide unique solutions particularly when dealing with nonlinear or complex equations.
By applying the boundary value, it restricts the infinite possibilities of solutions that the nonlinear differential equation might have. Despite not yielding to an explicit solution due to the complex integral involved, the initial value allowed for an implicit form.
IVPs are integral in real-world applications as they model conditions based on initial time or situations, ensuring results that match the practical scenarios.