Question

Solve the given initial value problem. \(y^{\prime}-y=\int_{0}^{t}(t-\lambda) e^{\lambda} d \lambda, \quad y(0)=-1\)

Short answer

Based on the initial value problem (IVP) involving a first-order differential equation, \(y'(t) - y(t) = \int_{0}^{t} (t-\lambda)e^{\lambda}d\lambda\) with the initial condition \(y(0) = -1\), we found the function \(y(t)\) to be: \(y(t) = e^t(t - 2)\)

Step-by-step solution

Evaluate the integral

We first want to compute the integral: \(\int_{0}^{t} (t-\lambda)e^{\lambda}d\lambda\) Notice that we can treat \(t\) as a constant since we're integrating with respect to \(\lambda\). We can rewrite this integral as the sum of two integrals: \(\int_{0}^{t} (t-\lambda)e^{\lambda}d\lambda = t\int_{0}^{t} e^{\lambda}d\lambda - \int_{0}^{t} \lambda e^{\lambda}d\lambda\) The first integral, \(\int_{0}^{t} e^{\lambda}d\lambda\), is straightforward to compute. The second one, \(\int_{0}^{t} \lambda e^{\lambda}d\lambda\), requires integration by parts. Let \(u = \lambda\) and \(dv = e^{\lambda}d\lambda\). Then, we have \(du = d\lambda\) and \(v= e^{\lambda}\). Using the integration by parts formula (\(\int u dv = uv - \int v du\)), we get: \(\int_{0}^{t} \lambda e^{\lambda}d\lambda = \lambda e^{\lambda} |_{0}^{t} - \int_{0}^{t} e^{\lambda}d\lambda\) Now, we can find the final integral: \(t\int_{0}^{t} e^{\lambda}d\lambda - \int_{0}^{t} \lambda e^{\lambda}d\lambda = te^{\lambda}|_{0}^{t} - (\lambda e^{\lambda} - e^{\lambda})|_{0}^{t}\)

Solve the differential equation using an integrating factor

The given differential equation is: \(y'(t) - y(t) = te^t - (t e^{t} - e^{t})\) Which can be simplified to: \(y'(t) - y(t) = e^t\) Now, we will use an integrating factor to solve this linear ODE. The integrating factor is given by \(I(t) = e^{\int -1 dt} := e^{-t}\). To apply the integrating factor, multiply both sides of the equation by \(e^{-t}\): \(e^{-t}y'(t) - e^{-t}y(t) = e^{-t}e^t\) And, the left side of the equation becomes the derivative of the product \(y(t)e^{-t}\) which is: \(\frac{d}{dt}(y(t)e^{-t}) = 1\)

Integrate and apply the initial condition

Now we can integrate both sides of this last equation with respect to \(t\): \(\int \frac{d}{dt}(y(t)e^{-t}) dt = \int dt\) Doing the integration, we get: \(y(t)e^{-t} = t + C\) Next, apply the initial condition \(y(0) = -1\). This gives us: \(-1 = e^0 + C \implies C = -2\)

Solve for \(y(t)\)

Now that we have the value of the constant \(C\), we can finally solve for \(y(t)\): \(y(t)e^{-t} = t - 2\) \(y(t) = e^t(t - 2)\) Therefore, the solution to the given initial value problem is: \(y(t) = e^t(t - 2)\)

Key concepts

Initial Value Problems

An initial value problem is a type of differential equation that comes with specific conditions, called initial conditions. These conditions are used to find a unique solution to the differential equation. In our problem, we're given the differential equation \(y'(t) - y(t) = \int_{0}^{t}(t-\lambda) e^{\lambda} d\lambda\) with an initial condition \(y(0)=-1\). The goal is to determine a function \(y(t)\) that satisfies both the differential equation and the initial condition.

• The initial condition is often provided as a specific value that the function and possibly its derivatives take at a particular point.
• In practical terms, the initial condition "sets the scene" for the solution. It helps narrow down the infinite possible solutions to one specific solution.

By solving the initial value problem, we determine how the function \(y(t)\) evolves from the initial condition across time or space.

Integration by Parts

Integration by parts is a technique used to solve integrals that are products of functions, like \(\int \lambda e^{\lambda} d\lambda\). It's based on the product rule for differentiation. The formula for integration by parts is \(\int u \, dv = uv - \int v \, du\). This swap simplifies the original integral into something more manageable.
For our exercise, we've decided:

• \(u = \lambda\)
• \(dv = e^{\lambda} d\lambda\)
• Therefore, \(du = d\lambda\) and \(v = e^{\lambda}\)

Applying integration by parts, we'll rework \(\int\lambda e^{\lambda} d\lambda\) into expressions that are easier to compute. This results in \(\lambda e^{\lambda}|_{0}^{t} - \int_{0}^{t} e^{\lambda} d\lambda\). The technique reduces a complicated integral into simpler parts that we can solve.

Linear Ordinary Differential Equations

Linear ordinary differential equations (ODEs) are equations involving derivatives that are linear in the unknown function and its derivatives. Our differential equation, \(y'(t) - y(t) = e^t\), is a first-order linear ODE.
Here are some key points about linear ODEs:

• They have a standard form: \(y' + p(t)y = g(t)\), where \(p(t)\) and \(g(t)\) are known functions.
• The solutions of linear ODEs are typically easier to find compared to nonlinear ones, especially using methods like integrating factors.
• The solution involves finding an expression for \(y(t)\) that satisfies the ODE across all values of \(t\) within a certain range.

Linear ODEs enable us to explore the behavior of dynamic systems under specific conditions.

Integrating Factors

Integrating factors are a powerful method to solve linear ordinary differential equations. The key idea is to multiply the entire differential equation by a cleverly chosen function, known as the integrating factor, which simplifies the process of integration.
In this exercise, we use \(I(t) = e^{\int -1 dt} = e^{-t}\) as the integrating factor. This choice helps convert the variable coefficient problem into a simpler equation by turning \(y'(t) - y(t)\) into the derivative of the product:

• Once multiplied, the left side becomes \(\frac{d}{dt}(y(t)e^{-t})\).
• This transformation simplifies the equation to \(1\), making it straightforward to integrate both sides.

The result of integrating both sides reveals a function that contains all potential solutions of the ODE. Finally, applying the initial condition lets us solve for any constants and find the specific solution, hence solving the problem.