Question

Find a solution to the initial value problem that is continuous on the given interval \([a, b]\). $$ y^{\prime}+p(t) y=2, \quad y(0)=1 ; \quad p(t)=\left\\{\begin{array}{lll} 0, & 0 \leq t \leq 1 & {[a, b]=[0,2]} \\ \frac{1}{t}, & 1

Short answer

\end{array}\right.$$ The solution to the initial value problem is given by the following piecewise-defined function: $$ y(t)=\left\\{\begin{array}{lll} 2t+1, & 0 \leq t \leq 1, \\\ t^2 + \frac{2}{t}, & 1 < t \leq 2. \end{array}\right. $$

Step-by-step solution

Find the integrating factor for each interval

First, we need to find the integrating factor for each interval of \(p(t)\). For the first interval \(0 \leq t \leq 1\), we have \(p(t) = 0\). For the second interval \(1 < t \leq 2\), we have \(p(t) = \frac{1}{t}\). The integrating factor for each interval is found by evaluating the exponential of the integral of \(p(t)\) in that interval. In our case, we have: 1. On the interval \(0 \leq t \leq 1\), \(P(t)=0\): $$ \text{Integrating factor} = e^{\int 0 dt } = e^0 = 1. $$ 2. On the interval \(1 < t \leq 2\), \(P(t)=\frac{1}{t}\): $$ \text{Integrating factor} = e^{\int\frac{1}{t} dt} = e^{\ln |t|} = t. $$

Solve the first-order linear DE on each interval separately

Now, we will multiply the whole equation by the integrating factor on each interval and then integrate to find the solution \(y(t)\). 1. On the interval \(0 \leq t \leq 1\), the integrating factor is \(1\), so the equation becomes: $$ y^{\prime}+0\cdot y = 2. $$ Integrate both sides with respect to \(t\): $$ \int(y^{\prime}) dt = \int 2 dt. $$ Now we get: $$ y(t)=2t+C_1 \quad \text{for} \quad 0 \leq t \leq 1. $$ 2. On the interval \(1 < t \leq 2\), the integrating factor is \(t\), so the equation becomes: $$ (ty^{\prime}+\frac{1}{t}\cdot t y) = 2t. $$ Integrate both sides with respect to \(t\): $$ \int(t y^{\prime}+y) dt = \int 2t dt. $$ Now we get: $$ ty(t) = t^3 + C_2 \quad \text{for} \quad 1 < t \leq 2. $$

Find the constants \(C_1\) and \(C_2\) using the initial condition and continuity

We have the initial condition \(y(0) = 1\). Apply this condition to the solution on the first interval, we get: $$ y(0) = 2 \cdot 0 + C_1 = 1 \implies C_1 = 1. $$ Now we have \(y(t) = 2t + 1\) for \(0 \leq t \leq 1\). Since we need a continuous solution over the whole interval \([0, 2]\), then values of \(y(t)\) must agree at the point \(t=1\), which implies: $$ 2 \cdot 1 + 1 = t y(1) \implies 3 = y_1(1). $$ Divide both sides by \(1\) and calculate \(C_2\) using the above equation: $$ y_1(1) = (1)^3 + C_2 = 3 \implies C_2 = 2. $$ Now we have \(ty(t) = t^3 + 2\) for \(1 < t \leq 2\), or equivalently, \(y(t) = t^2 + \frac{2}{t}\) for \(1 < t \leq 2\).

Write down the final piecewise-defined solution

So, the solution to the initial value problem is given by the following piecewise-defined function: $$ y(t)=\left\\{\begin{array}{lll} 2t+1, & 0 \leq t \leq 1, \\\ t^2 + \frac{2}{t}, & 1<t \leq 2. \end{array}\right. $$ This function is continuous on the interval \([0, 2]\) and satisfies the initial value problem.

Key concepts

Integrating Factor

An integrating factor is a powerful tool used in solving first-order linear differential equations. It allows us to transform a non-exact differential equation into an exact one, making it easier to solve. The integrating factor is often a function, commonly denoted as \( ext{IF}(t) \), which, when multiplied by the entire differential equation, simplifies the process of integration.

To find the integrating factor for an equation of the form \( y' + p(t)y = g(t) \), compute the exponential of the integral of the function \( p(t) \):

• Calculate \( e^{\int p(t) \, dt} \).

For example:

• If \( p(t) = 0 \), the integrating factor becomes \( e^{0} = 1 \).
• If \( p(t) = \frac{1}{t} \), the integrating factor is \( t \) because \( e^{\ln |t|} = t \).

By applying this factor, we can often integrate both sides of the equation more easily, leading toward the solution.

First-order Linear Differential Equation

First-order linear differential equations are a type of equation involving derivatives, where the highest derivative is first order. These equations typically have the standard form \( y' + p(t)y = g(t) \), representing a straight line relationship when plotted.

The steps to solve a first-order linear differential equation usually involve:

• Identifying the functions \( p(t) \) and \( g(t) \).
• Finding the integrating factor as described above.
• Multiplying through by the integrating factor to facilitate integration.

These equations are prevalent in various applications, ranging from physics to finance, wherever systems change at rates proportional to their own value or some external function.

Continuous Solution

A continuous solution of a differential equation implies that the solution is smooth and uninterrupted over the interval of interest, which, in this problem, is \([0, 2]\). This means that the function \( y(t) \) does not have any breaks, jumps, or holes across the interval.

To ensure continuity, particularly when dealing with piecewise functions, we verify that the function values match at the boundary points between pieces. For instance, between the intervals \([0, 1]\) and \((1, 2]\), it’s crucial that the solutions on both intervals converge at \( t = 1 \). In the given problem, solving the differential equations separately for \([0, 1]\) and \((1, 2]\) ensures that:

• The value from \( y_1(t) = 2t + 1 \) at \( t = 1 \) is equal to the value from \( y_2(t) = t^2 + \frac{2}{t} \) at \( t = 1 \).

This agreement guarantees that \( y(t) \) is a continuous solution across the entire interval.

Piecewise Function

A piecewise function is one that is defined by different expressions over disjoint intervals. This means the function can exhibit different behaviors while maintaining a continuous nature across the specified domain.

In the context of differential equations, especially initial value problems like the one here, using a piecewise function is helpful in describing solutions where the conditions or equations change at certain points. For our problem, the function \( y(t) \) is defined differently on the intervals \([0, 1]\) and \((1, 2]\).

The solution is given by:

• \( y(t) = 2t + 1 \) for \( 0 \leq t \leq 1 \).
• \( y(t) = t^2 + \frac{2}{t} \) for \( 1 < t \leq 2 \).

When assembled, these intervals form a continuous, complete solution over their entire range, satisfying the initial condition as well as the piecewise constraints.