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. $$ \frac{d y}{d t}+e^{y} t=e^{y} \sin t, \quad y(0)=0 $$

Short answer

Question: Find the interval of existence of the solution of the initial value problem \( \frac{dy}{dt} + e^y t = e^y \sin t\), \(y(0)=0\) Answer: Since we could not find an explicit solution, the interval of existence cannot be determined for the given initial value problem. The solution is implicitly given in terms of \(y\) and \(t\).

Step-by-step solution

Identify the type of differential equation

The given differential equation is of the form \( \frac{dy}{dt} + e^y t = e^y \sin t\) along with the initial value \(y(0)=0\). First, we need to examine the equation to establish that it's a first-order linear differential equation.

Obtain an implicit solution

To find the implicit solution, we'll manipulate the equation to put it in standard form for a first-order linear differential equation: $$ \frac{d y}{d t}+e^{y} t - e^y\sin t = 0$$ Multiply by integrating factor \(e^y\): $$ e^y \frac{d y}{d t} + e^{2y} t - e^{2y} \sin t = 0 $$ Observe that the left-hand side is equal to the derivative of \((e^yt)\) with respect to \(t\): $$ \frac{d}{dt}(e^y t) = e^{2y} \sin t$$ Integrate both sides with respect to \(t\): $$ e^y t = \int e^{2y}\sin{t} \, dt + C $$ That's our implicit solution.

Look for an explicit solution if possible

An explicit solution would be completing the following procedure: $$ y = f(t) $$ In this case, it's nearly impossible to find an explicit function \(f(t)\) that exactly gives the relationship between \(y\) and \(t\). Therefore, we only have an implicit solution, as found above.

Determine the t-interval of existence

Since we were unable to find an explicit solution, we cannot determine the \(t\)-interval of existence. The solution is only implicitly given in terms of \(y\) and \(t\).

Key concepts

Implicit Solution

An implicit solution to a differential equation is a type of solution where the dependent variable, usually expressed as a function of the independent variable, is not isolated on one side of the equation. It involves both variables together in a single equation without solving for one specific outcome. In this particular problem, the differential equation given is \[ \frac{d y}{d t} + e^y t = e^y \sin t \]With manipulation, it was transformed into:\[ e^y t = \int e^{2y}\sin{t} \, dt + C \]Here, we see both variables, \( y \) and \( t \), are present together and not separated into a simple function like \( y = f(t) \). This expression is what is called an implicit solution. It implies a relationship between \( y \) and \( t \) without explicitly solving for \( y \) in terms of \( t \).
Implicit solutions are common when differential equations are complex. They provide us with valuable relationships and can sometimes be sufficient depending on the context of the problem.

Explicit Solution

An explicit solution is one where the dependent variable is fully disentangled and expressed solely in terms of the independent variable. The main goal is to express the variable \( y \) purely as a function of \( t \), denoted as \( y = f(t) \). Finding such solutions makes it easier to understand how changes in the independent variable affect the dependent one.In this exercise, while we sought an explicit solution, we determined it's practically challenging due to the equation's intricacies. The equation after integration and manipulation didn't readily yield a straightforward \( y = f(t) \). This doesn't mean explicit solutions are impossible for all problems, but in cases like this, they might require advanced methods or may not be possible analytically at all. Implicit solutions can often provide adequate information, but if an explicit solution is available, it typically offers clearer insights.

Integrating Factor

An integrating factor is a crucial concept in solving linear first-order differential equations. It is essentially a function that, when multiplied by the original differential equation, transforms it into a simpler form, often one that is easily integrable.In our example, the differential equation:\[ \frac{d y}{d t} + e^y t = e^y \sin t \]required manipulation. By multiplying through by the integrating factor \( e^y \), the left side of the equation forms the product rule derivative:\[ e^y \frac{d y}{d t} + e^{2y} t - e^{2y} \sin t = 0 \]This manipulation is designed to incorporate the original differential terms into a format that reveals its integral structure, allowing easy integration and thus simplification.Using integrating factors is a powerful technique that is instrumental when dealing with first-order linear differential equations and transforms seemingly complex equations into something much more manageable.

Initial Value Problem

An initial value problem (IVP) involves finding a solution to a differential equation that meets specified initial conditions. These are important to ensure that the solution is unique and applicable to real-world situations.In our exercise, the initial condition provided was \( y(0) = 0 \). This means that when \( t = 0 \), the value of \( y \) must be 0. Such conditions are vital in solving differential equations as they help in identifying one specific solution from potentially many.The role of initial conditions comes after solving the general differential equation, aligning the solution with the real-world scenario presented. Without these conditions, the solutions to differential equations could be less meaningful practically, representing a whole family of potential results rather than one precise answer.