Question

\(y^{\prime}=-y+e^{t} y^{2}, \quad y(-1)=-1\)

Short answer

Question: Solve the first-order non-linear differential equation with initial condition \(y' + y = e^t y^2\) and \(y(-1)=-1\). Answer: The solution to the given non-linear differential equation with the initial condition is expressed implicitly as: \(y = \frac{1}{e^t} \left(\int e^{2t} y^2 dt - e\right)\) Note that a more explicit solution cannot be determined in terms of elementary functions.

Step-by-step solution

Rewrite the equation

We rewrite the given differential equation as: \(y' + y = e^t y^2\)

Find an integrating factor

Recall that an integrating factor for a first-order linear differential equation in the form \(y' + P(t)y = Q(t)\) is given by: \(I(t) = e^{\int P(t) dt}\) In this case, we have \(P(t) = 1\), so the integrating factor is given by: \(I(t) = e^{\int 1 dt} = e^t\)

Multiply both sides by the integrating factor

We multiply both sides of the equation by the integrating factor \(e^t\): \(e^t y' + e^t y = e^{2t} y^2\) Notice that the left side of the equation can be written as the derivative of the product \(y(t)e^t\): \(\frac{d}{dt}(ye^t) = e^{2t} y^2\)

Integrate both sides

Now we integrate both sides of the equation with respect to \(t\): \(\int \frac{d}{dt}(ye^t) dt = \int e^{2t} y^2 dt\) After integrating, we get: \(ye^t = \int e^{2t} y^2 dt + C\)

Solve for \(y\)

To solve for \(y\), we divide both sides of the equation by \(e^t\): \(y = \frac{1}{e^t} \left( \int e^{2t} y^2 dt + C\right)\) At this point, we can't proceed further without knowing more information, as this equation has no general solution in terms of elementary functions. However, we can still apply the initial condition.

Apply the initial condition

We are given that \(y(-1)=-1\). We substitute this into the equation we derived: \(-1 = \frac{1}{e^{-1}} \left(\int e^{2(-1)} (-1)^2 dt + C\right)\) Solving for \(C\), we get: \(-e = \int e^{-2} dt + C\) This equation can't be solved further in terms of elementary functions, but we have expressed the solution for the given non-linear differential equation in an implicit form, including the initial condition, as: \(y = \frac{1}{e^t} \left(\int e^{2t} y^2 dt - e\right)\)

Key concepts

Integrating Factor

The concept of an integrating factor is a crucial tool for solving first-order linear differential equations. It is essentially a function, typically denoted by \( I(t) \), that when multiplied by the given differential equation, simplifies it into an easily integrable form. This process helps in transforming the equation so that the left-hand side becomes the derivative of a product, which can then be integrated with respect to \( t \).

For the equation \( y' + P(t)y = Q(t) \), the integrating factor is determined by taking the exponential of the integral of \( P(t) \), written mathematically as \( I(t) = e^{\int P(t) dt} \). In our specific exercise, with \( P(t) = 1 \), the integrating factor simplifies to \( e^t \). Upon multiplication, this has the wonderful property of making the left side of the differential equation a perfect derivative, which sets the stage for straightforward integration and eventually solving for \( y(t) \).

First-Order Linear Differential Equation

First-order linear differential equations are equations of the form \( y' + P(t)y = Q(t) \), where \( y' \) denotes the derivative of \( y \) with respect to \( t \), and \( P(t) \) and \( Q(t) \) are functions of \( t \). They are called 'linear' because the dependent variable \( y \) and its derivative appear to the first power and not as products or exponents.

These equations crop up in numerous scientific and engineering fields, reflecting changes in systems over time. Our exercise features such an equation that has been cleverly manipulated into the correct linear form. Solving these equations typically involves finding an integrating factor to facilitate the integration process, which is a systematic approach to finding solutions that describe the behavior of dynamic systems.

Initial Value Problem

An initial value problem is a differential equation accompanied by a specific value, called the initial condition, which the solution must satisfy at a particular point. For our exercise, the initial value problem is given by \( y(-1) = -1 \), where the solution to the differential equation must pass through the point \( (-1, -1) \) on the \( ty \)-plane.

These problems are essential in predicting the future behavior of dynamical systems, ensuring that the solutions are unique and applicable to real-world scenarios. When solving an initial value problem, we use the initial condition to determine any constants that arise from the integration process, ultimately providing a particular solution to the equation that fits the initial scenario perfectly.

Implicit Solution

An implicit solution to a differential equation is an expression that defines the dependent variable \( y \) in terms of the independent variable \( t \) indirectly rather than providing a direct definition as in explicit solutions. It often includes an integral or other terms which are left in terms of \( y \) itself.

In many cases, like the one in our textbook exercise, an explicit, closed-form solution may not be achievable. In such instances, the implicit solution becomes a powerful way to characterize the relationship between variables, even if it doesn't give us a direct formula. While it may seem abstract, implicit solutions can still provide critical insights into the system being modeled, and in some cases, numerical methods can be used to approximate values based on the implicit solution.