Question

Consider the initial value problem $$ y^{\prime}+\frac{1}{4} y=3+2 \cos 2 t, \quad y(0)=0 $$ (a) Find the solution of this initial value problem and describe its bchavior for large \(t\). (b) Determine the value of \(t\) for which the solution first intersects the line \(y=12\).

Short answer

A: The integrating factor is \(I(t) = e^{\frac{1}{4}t}\). Q: What is the solution to the given equation with the initial condition? A: The solution is \(y(t) = e^{-\frac{1}{4}t}(12 - 4\cos{2t} - 8)\). Q: As t approaches infinity, what does the solution converge to? A: As \(t \rightarrow \infty\), the solution converges to 0. Q: When does the solution first intersect the line y=12? A: The solution first intersects the line \(y=12\) at \(t=\frac{3\pi}{4}\).

Step-by-step solution

To find the integrating factor, recall the general form of a first-order linear inhomogeneous ODE: $$ y' + p(t)y = g(t) $$ In our case, we have \(p(t) = \frac{1}{4}\) and \(g(t) = 3 + 2 \cos{2t}\). The integrating factor is given by $$ I(t) = e^{\int p(t)dt} = e^{\int \frac{1}{4} dt} = e^{\frac{1}{4}t} $$ #Step 2: Multiply the given ODE by the integrating factor#

We multiply the given equation by the integrating factor \(I(t) = e^{\frac{1}{4}t}\): $$ (e^{\frac{1}{4}t}y') + \frac{1}{4}e^{\frac{1}{4}t}y = (3+2\cos{2t})e^{\frac{1}{4}t} $$ Notice that the left-hand side is the derivative of the product \((e^{\frac{1}{4}t} y)\). #Step 3: Integrate both sides of the equation#

Since the left-hand side is the derivative \((e^{\frac{1}{4}t}y)'\), integrate both sides with respect to \(t\): $$ \int\! (e^{\frac{1}{4}t}y)' dt = \int\! (3+2\cos{2t})e^{\frac{1}{4}t} dt $$ Integration gives: $$ e^{\frac{1}{4}t}y = 12e^{\frac{1}{4}t} - 4e^{\frac{1}{4}t}\cos{2t} + C $$ #Step 4: Use the initial condition to find the constant of integration#

We are given the initial condition \(y(0) = 0\). Substitute \(t=0\) and \(y(0)=0\) in the equation: $$ e^{0}y(0) = 12e^0 - 4e^0\cos{0} + C $$ This gives: $$ 0 = 12 - 4 + C \Rightarrow C = -8 $$ #Step 5: Solve for y(t) and analyze its behavior for large t#

Now we can write the solution as follows: $$ y(t) = e^{-\frac{1}{4}t}(12 - 4\cos{2t} - 8) $$ For large \(t\), the exponential term \(e^{-\frac{1}{4}t}\) tends to zero. Therefore, as \(t \rightarrow \infty\), the solution converges to 0. #Step 6: Determine the value of t when the solution first intersects y=12#

We need to find the value of t when our solution intersects the line y=12. Set \(y(t)=12\) and solve for \(t\): $$ 12 = e^{-\frac{1}{4}t}(12 - 4\cos{2t} - 8) $$ Simplifying, we get: $$ 1 = e^{-\frac{1}{4}t}(1 - 2\cos{2t}) $$ Before proceeding further, we will analyze the initial condition of the cosine function. Given that, \(y(0) = 0\), we have \({1 - 2\cos{2(0)}} < 0\). To make \(y(t)\) positive, we look for the value of \(t\) in the range \((0, \pi)\) such that cosine is negative. Thus, we have: $$ 2t = \frac{3\pi}{2} \Rightarrow t = \frac{3\pi}{4} $$ As a result, the solution \(y(t)\) first intersects the line \(y=12\) at \(t=\frac{3\pi}{4}\).

Key concepts

Differential Equations

Differential equations are a cornerstone of mathematics and are essential in modeling the behavior of various natural phenomena. Fundamentally, they are equations that relate a function with its derivatives. In our given exercise, we encounter a first-order linear ordinary differential equation (ODE), which is among the simplest and most commonly studied types.

In such differential equations, the solution y(t) represents a function that, if differentiated and substituted back into the equation along with any specified terms, makes the equation true for all values of t. The specific problem we're looking at involves finding the function y(t) that satisfies both the equation and the initial condition y(0)=0.

Integrating Factor

The integrating factor is a strategic approach to solve certain types of ODEs, specifically linear equations. It is a function, which, when multiplied to the entire equation, transforms it into an easily integrable form. This method is specifically clever in that it turns the left-hand side of the ODE into the derivative of a product, which is straightforward to integrate.

To apply this method, one must first identify the function p(t) in the standard form of the equation, and then calculate the integrating factor as \(I(t) = e^{\int p(t) \, dt}\). In our exercise, by multiplying the original equation by the integrating factor, we are able to integrate and solve for y(t). This is an invaluable technique for solving ODEs that, while may seem complex, neatly unpacks the problem into a solvable format.

Boundary Value

Boundary values play a pivotal role in solving differential equations since they provide necessary conditions that the solution must satisfy. In initial value problems, like the one featured in our exercise, the 'initial condition' provides a boundary value at a particular point in time (in this case, t=0), giving us a starting point for the solution.

This constraint is crucial as it allows us to find a specific solution to the ODE that not only meets the mathematical specifications of the equation but also aligns with the physical or abstract scenario being modeled. It ensures the uniqueness of the solution which, without such a boundary condition, could be one of many possibilities. The boundary value basically 'guides' the solution to fit the initial scenario described by the problem.