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Magazine Chapter 5: Problem 28
Solve the given initial value problem. \(y^{\prime}-y=\int_{0}^{t}(t-\lambda) e^{\lambda} d \lambda, \quad y(0)=-1\)
Short Answer
Expert verified
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
01
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:
\(t\int_{0}^{t} e^{\lambda}d\lambda - \int_{0}^{t} \lambda e^{\lambda}d\lambda = te^{\lambda}|_{0}^{t} - (\lambda e^{\lambda} - e^{\lambda})|_{0}^{t}\)
02
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:
\(\frac{d}{dt}(y(t)e^{-t}) = 1\)
03
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:
\(-1 = e^0 + C \implies C = -2\)
04
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:
\(y(t) = e^t(t - 2)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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:
For our exercise, we've decided:
- \(u = \lambda\)
- \(dv = e^{\lambda} d\lambda\)
- Therefore, \(du = d\lambda\) and \(v = e^{\lambda}\)
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:
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.
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:
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.
