Question

\(y^{\prime}=2 t y(1-y), \quad y(0)=-1\)

Short answer

#Answer# The given initial value problem with the ODE \(y^{\prime}=2 t y(1-y), \quad y(0)=-1\) has no valid solutions.

Step-by-step solution

Find an integrating factor

An integrating factor is a function \(\mu(t)\) that simplifies our ODE. To find the integrating factor, we first need to rewrite the given ODE as a first-order linear ODE: \(y^{\prime} - 2ty(1-y) = 0\) In this case, we can notice that the linear ODE is already in standard form: \(y^{\prime} + P(t)y = Q(t)\) Where \(P(t) = -2t(1-y)\) and \(Q(t) = 0\). Now we can find the integrating factor, \(\mu(t)\): \(\mu(t) = e^{\int P(t) dt} = e^{\int (-2t(1-y)) dt}\) Given that we don't know the value of 'y', let's rewrite the equation without 'y': \(\mu(t) = e^{(-t^2)}\)

Multiply the ODE by the integrating factor

Now, we will multiply the entire ODE by the integrating factor: \(e^{(-t^2)}(y^{\prime} - 2ty(1-y)) = 0 * e^{(-t^2)}\) This simplifies the ODE into: \(e^{(-t^2)}y^{\prime} - 2te^{(-t^2)}y(1-y) = 0\)

Integrate both sides

Now, integrate both sides with respect to 't': \(\int(e^{(-t^2)}y^{\prime}) dt - \int(2te^{(-t^2)}y(1-y)) dt = \int(0) dt\) Unfortunately, the integral of second term is quite complicated and cannot be solved using elementary functions. However, let's notice that our ODE is separable: \(y^{\prime}=2 t y(1-y)\) can be written as \(-\frac{dy}{1-y} = 2tdt\)

Integrate both sides (adjusted for a separable ODE)

Now, integrate both sides of the separable equation: \(\int -\frac{dy}{1-y} = \int 2t dt\) \(- \ln|1-y| = t^2 + C\) To solve for y, we first need to get rid of the natural logarithm. We can do that by taking the exponent of both sides: \(e^{- \ln|1-y|} = e^{t^2 + C}\) \(|1-y| = e^{t^2}e^{C}\)

Solve for the constant using the initial condition

Now, use the initial condition \(y(0)=-1\) to solve for the constant 'C': \(|-1-y(0)| = e^{0}e^{C}\) \(|-1+1| = e^{C}\) \(0 = e^{C}\) As this is not possible (exponential function never equals 0), there is no solution that satisfies the initial condition. In conclusion, the given initial value problem with the ODE \(y^{\prime}=2 t y(1-y), \quad y(0)=-1\) has no valid solutions.

Key concepts

Integrating Factor

An integrating factor is a mathematical tool used to transform a non-exact first-order linear ordinary differential equation (ODE) into an exact ODE, which is easier to solve. It is often denoted as \(\mu(t)\) and is derived from the standard form of a first-order linear ODE, which is \(y' + P(t)y = Q(t)\). The integrating factor is calculated as \(\mu(t) = e^{\int P(t) dt}\).

Now, why is it called an 'integrating factor'? Essentially, once you multiply the entire ODE by \(\mu(t)\), the left-hand side of the equation becomes the derivative with respect to 't' of the product \(\mu(t)y\). This converted form can then be integrated with respect to 't', leading neatly to a solution. But, remember, choosing an integrating factor does not guarantee a straightforward integral; for some functions, the integral may still be difficult or even impossible to solve analytically, which is precisely what happened in the presented exercise.

Separable ODE

A separable ordinary differential equation is one where the variables can be separated on different sides of the equation, allowing the equation to be written in the form \(f(y)dy = g(t)dt\). This is helpful because it allows us to integrate each side of the equation with respect to its own variable.

Once the variables are separated, as seen in the exercise above, the integration process often becomes much simpler compared to dealing with the paired variables together. The process involves finding the antiderivatives of each side, leading to an implicit solution involving 'y' and 't'. Remember that additional steps are usually required to isolate 'y' and to apply any initial conditions.

Initial Value Problem

An initial value problem comprises a differential equation coupled with a specific requirement that the solution must satisfy an initial condition, usually given in the form \(y(t_{0}) = y_{0}\). This condition anchors the solution to a specific point and ensures a unique solution for differential equations satisfying certain conditions, known as existence and uniqueness theorems.

In the exercise, the initial value problem was used in an attempt to determine the constant of integration. However, the given initial condition led to an inconsistency, meaning that within the context of real-valued functions, no solution exists that can satisfy the initial condition \(y(0) = -1\). This highlights the importance of initial condition compatibility in the search for solutions to differential equations.

First-order Linear ODE

A first-order linear ODE is an equation of the form \(y' + p(t)y = q(t)\), where \(y\) is the unknown function of \(t\), and \(p(t)\) and \(q(t)\) are known functions of \(t\). It is called 'first-order' because it involves only the first derivative of \(y\), and 'linear' because \(y\) and its derivative appear to the first power and don't multiply with each other.

These equations are essential since they commonly arise in physical systems and mathematical modeling of numerous phenomena. They have a well-defined procedure for finding solutions, often entailing the use of integrating factors when \(q(t)\) is non-zero. However, the example from the exercise interestingly revealed a situation where, despite the equation being a first-order linear ODE, the initial condition presented led to an unsolvable scenario, illustrating the diverse and sometimes unexpected nature of solving differential equations.