Question

Solving initial value problems Determine whether the following equations are separable. If so, solve the initial value problem. $$\frac{d y}{d t}=t y+2, y(1)=2$$

Short answer

Initial Problem: $$\frac{d y}{dt} = ty + 2, \quad y(1) = 2$$ Answer: The given first-order differential equation is not separable. An alternative method to solve the problem is using integrating factors.

Step-by-step solution

Write the given equation and initial condition

The given differential equation and initial condition are: $$\frac{d y}{dt} = ty + 2, \quad y(1) = 2$$

Check if the equation is separable

We want to determine whether the given equation can be written as \(\frac{dy}{g(y)} = f(t) \, dt\). In this case, we can rearrange the terms: $$\frac{d y}{dt} - ty = 2$$ Unfortunately, we cannot separate the variables, so the equation is not separable.

Trying an alternative approach using integrating factors

Though the given equation is not separable, we can try solving it using the method of integrating factors. To do this, we write the equation as a linear first-order differential equation in the form: $$\frac{dy}{dt} - ty = 2$$ The integrating factor, \(\mu(t)\), is computed by finding the exponential of the integral of the function multiplying the dependent variable, in this case, \(-t\). So we have: $$\mu(t) = e^{\int{-t \, dt}} = e^{-\frac{1}{2}t^2}$$ Now, we multiply the entire differential equation by the integrating factor: $$e^{-\frac{1}{2}t^2}\frac{dy}{dt} - te^{-\frac{1}{2}t^2}y = 2e^{-\frac{1}{2}t^2}$$ By doing this, the left side of the equation is now the derivative of the product of y(t) and the integrating factor, which is: $$\frac{d}{dt}(y(t)e^{-\frac{1}{2}t^2}) = 2e^{-\frac{1}{2}t^2}$$

Integrate both sides of the equation

Now we integrate both sides of the equation with respect to t: $$\int\frac{d}{dt}(y(t)e^{-\frac{1}{2}t^2})\, dt = \int{2e^{-\frac{1}{2}t^2}\, dt}$$ This gives us: $$y(t)e^{-\frac{1}{2}t^2} = \int{2e^{-\frac{1}{2}t^2}\, dt} + C$$

Solve for y(t)

Now, we solve for y(t) by multiplying both sides by \(e^{\frac{1}{2}t^2}\): $$y(t) = e^{\frac{1}{2}t^2}\left(\int{2e^{-\frac{1}{2}t^2}\, dt} + C\right)$$ Unfortunately, the integral \(\int{2e^{-\frac{1}{2}t^2}\, dt}\) does not have a simple closed-form solution. However, we can leave it in its current form since the exercise only requires determining whether the equation is separable, and we have shown that it is not.

Key concepts

Separable Differential Equation

Understanding separable differential equations is essential when dealing with certain types of first-order differential equations. A differential equation is considered separable if it can be written in the form \( \frac{dy}{dx} = g(x)h(y) \) where the functions \( g(x) \) and \( h(y) \) contain only \( x \) and \( y \) terms respectively. This allows the variables to be separated onto different sides of the equation.

To solve a separable differential equation, one typically follows these steps:

• Rearrange the equation to isolate \( dx \) terms on one side and \( dy \) terms on the other.
• Integrate both sides independently.
• Find the constant of integration, if necessary, using any given initial conditions.

In the case of the exercise provided, the equation \( \frac{d y}{d t} = t y + 2 \) cannot be rearranged to separate the variables, and thus does not fit the definition of a separable differential equation.

Integrating Factor

An integrating factor is a function that is used to simplify the process of solving linear first-order differential equations that are not separable. The integrating factor, typically denoted as \( \mu(t) \) or \( \mu(x) \) depending on the variable, is calculated by taking the exponential of the integral of the coefficient of \( y \) in the differential equation.

For the differential equation \( \frac{dy}{dt} - ty = 2 \) from our exercise, the integrating factor is computed as \( \mu(t) = e^{\int{-t \, dt}} = e^{-\frac{1}{2}t^2} \). By multiplying every term of the equation by this integrating factor, the left-hand side becomes the derivative of the product of \( y(t) \) and \( \mu(t) \), simplifying the equation into a form that can be integrated with respect to \( t \) directly.

The general steps to use an integrating factor are:

• Identify the appropriate form \( \frac{dy}{dx} + P(x)y = Q(x) \) and calculate \( \mu(x) \) or \( \mu(t) \) accordingly.
• Multiply through by \( \mu(x) \) (or \( \mu(t) \)) to simplify the equation.
• Integrate both sides, often leading to an easier integration.
• Find the constant of integration using any given initial conditions, if needed.

Linear First-Order Differential Equation

A linear first-order differential equation has the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \) only. These equations feature a first derivative of the unknown function \( y \) and the function \( y \) itself, both linearly without any powers or products of \( y \) and \( \frac{dy}{dx} \).

To tackle such equations, the approach involving an integrating factor is often employed, especially when they are non-separable as seen in the provided exercise. The key characteristic of these equations is their first-order nature, meaning they involve only the first derivative of \( y \), and linearity, the absence of non-linear terms like \( y^2 \) or \( sin(y) \).

When solving linear first-order differential equations, the integrating factor method transforms them into an exact differential form, which can be integrated easily. This method is crucial and often included in the curriculum of differential equations because it provides a systematic way of finding solutions to a common class of differential equations that would otherwise be difficult to solve.