Question

A special class of first-order linear equations have the form \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=f(t),\) where \(a\) and \(f\) are given functions of t. Notice that the left side of this equation can be written as the derivative of a product, so the equation has the form $$a(t) y^{\prime}(t)+a^{\prime}(t) y(t)=\frac{d}{d t}(a(t) y(t))=f(t)$$ Therefore, the equation can be solved by integrating both sides with respect to \(t .\) Use this idea to solve the following initial value problems. $$t^{3} y^{\prime}(t)+3 t^{2} y=\frac{1+t}{t}, y(1)=6$$

Short answer

Question: Solve the initial value problem: $$t^{3}y'(t)+3t^{2}y(t)=\frac{1+t}{t}, \quad y(1) = 6$$ Answer: The solution to the initial value problem is: $$y(t) = \frac{\ln|t| + t + 5}{t^3}$$

Step-by-step solution

Identifying Functions

First, let's identify the functions \(a(t)\), \(a'(t)\) and \(f(t)\) in the given equation: $$t^{3}y'(t)+3t^{2}y(t)=\frac{1+t}{t}$$ We can see that \(a(t) = t^3\), and therefore \(a'(t) = 3t^2\). The function \(f(t)\) is given by \(\frac{1+t}{t}\).

Verify the Equation Form

To confirm that our equation falls into the special class mentioned, we can check if \(a(t) y^{\prime}(t)+a^{\prime}(t) y(t)\) is equal to \(\frac{d}{dt}(a(t)y(t))\): $$\frac{d}{dt}(t^3y(t)) = (t^3)'y(t) + t^3y'(t) = 3t^2y(t) + t^3y'(t)$$ As we can see, this matches the left side of our given equation, so we can proceed with the method described.

Integrating Both Sides

Now we will integrate both sides of the equation with respect to \(t\): $$\int \frac{d}{dt}(t^3y(t)) dt = \int \frac{1+t}{t} dt$$ The left side simplifies to: $$\int d(t^3y(t)) = t^3y(t)$$ For the right side, we can split the integrand into two parts: $$\int \frac{1+t}{t} dt = \int \frac{1}{t} dt + \int 1 dt$$ Solving the integrals on the right side: $$\int \frac{1}{t} dt = \ln|t| + C_1$$ $$\int 1 dt = t + C_2$$ Adding the constants together, we get: $$t^3y(t) = \ln|t| + t + C$$

Solving for y(t)

Now, we will solve for \(y(t)\): $$y(t) = \frac{\ln|t| + t + C}{t^3}$$

Utilizing the Initial Condition

Next, we will use the initial condition \(y(1) = 6\) to find the value of the constant \(C\): $$6 = \frac{\ln|1| + 1 + C}{1^3}$$ Simplifying this equation gives us \(C = 5\). Now we can write the particular solution to the initial value problem: $$y(t) = \frac{\ln|t|+t+5}{t^3}$$ Therefore, the solution to the initial value problem is: $$y(t) = \frac{\ln|t| + t + 5}{t^3}$$

Key concepts

Integrating Factor

The integrating factor is a function used to solve linear differential equations of the form \(y' + p(t)y = g(t)\). If an equation isn't readily solvable, we multiply each term by an integrating factor, \(I(t)\), which is normally derived from the function \(p(t)\). Specifically, the integrating factor is given by \(e^{\int p(t) dt}\), and it transforms the equation into an exact differential, making it easier to solve. In the exercise provided, the integrating factor isn't required because the left side of the equation is already an exact derivative of the function \(a(t)y(t)\). However, in cases where an exact derivative isn't present, using the integrating factor would simplify the problem by turning the left side into a derivative of the product \(I(t)y(t)\), which can then be integrated to solve for \(y(t)\).

Initial Value Problem

An initial value problem is a differential equation accompanied by a specific value, known as the initial condition, at a given point. The aim is to find a specific solution that not only satisfies the differential equation but also fulfills the initial condition. In our exercise, \(y(1) = 6\) serves this purpose. This condition allows us to solve for the constant of integration, which arises when integrating functions. For our equation, after integrating both sides, we substituted \(t = 1\) and \(y = 6\) into the equation to solve for the constant \(C\). The inclusion of the initial condition transforms a general solution with an arbitrary constant into a unique, specific solution, tailored to the initial value provided.

Separable Differential Equations

Separable differential equations are those in which the variables can be separated on opposite sides of the equation, allowing us to integrate each side with respect to its corresponding variable. Typically, these equations are of the form \(y' = g(x)h(y)\), meaning we can isolate all \(y\)-terms on one side and all \(x\)-terms on the other. In our exercise, the equation is not separable since it is a first-order linear differential equation, not of the form that permits separation of variables. However, the concept of separable equations is fundamental in understanding how to approach different types of differential equations, and it is important to recognize when an equation can be separated to facilitate easier integration.

Integration of Functions

Integration is a core operation in calculus that is used to find areas, volumes, central points, and many useful things. In essence, integration reverses differentiation, finding the function when its rate of change is known. The integral of a function \(f(x)\) with respect to variable \(x\) is denoted by \(\int f(x) dx\) and represents the accumulation of the quantities of \(f\) as \(x\) changes. When we integrate to solve differential equations, we often encounter an arbitrary constant of integration (denoted as \(C\)), which represents all possible values of the indefinite integral. In the solution for the provided exercise, we performed integration to solve for \(y(t)\), considering both sides of the equation. It is the fundamental method by which we solve first-order linear differential equations when an exact derivative is present, as in this case scenario.