Question

Solve the initial value problem $$ y^{\prime}+2 y=g(t), \quad y(0)=0 $$ where $$ g(t)=\left\\{\begin{array}{ll}{1,} & {0 \leq t \leq 1} \\ {0,} & {t>1}\end{array}\right. $$

Short answer

Answer: The piecewise defined solution to the given initial value problem is $$ y(t) =\begin{cases} e^{-2t}\left(\int_0^t e^{2\tau}\,d\tau \right) & 0 \leq t \leq 1 \\ e^{-2t}-\frac{1}{e} & t > 1 \end{cases} $$

Step-by-step solution

Identify and multiply with integrating factor

To find the integrating factor, we first need to identify the coefficient of y in the given differential equation, which is 2. The integrating factor (IF) will be \(\displaystyle e^{\int 2dt}=e^{2t}\). Multiply the given equation by the integrating factor. $$ e^{2t}(y' + 2y) = e^{2t}g(t) $$

Integrate to find the general solution

Integrate both sides of the equation with respect to t. $$ \int e^{2t}(y' + 2y) \, dt = \int e^{2t}g(t) \, dt $$ The left-hand side simplifies due to the product rule of integration: $$ y e^{2t} = \int e^{2t}g(t) \, dt + C $$ Now, solve for y by dividing both sides by \(\displaystyle e^{2t}\): $$ y(t) = e^{-2t} \left(\int e^{2t}g(t) \, dt + C\right) $$

Solve for different intervals

We'll solve this equation for the different intervals given in the piecewise function for g(t). 1. For \(\displaystyle 0\leq t\leq 1\), we have \(\displaystyle g(t)=1\): $$ y(t) = e^{-2t}\left(\int e^{2t}(1)\,dt + C\right) $$ Let's call the function obtained in this interval as \(\displaystyle y_{1}( t)\) 2. For \(\displaystyle t > 1\), we have \(\displaystyle g(t) =0\) $$ y(t) = e^{-2t}\left(\int e^{2t}(0) \, dt + C \right) $$ Let's call the function obtained in this interval as \(\displaystyle y_{2}( t)\)

Find the constants using initial and continuity conditions

Using the initial condition \(\displaystyle y( 0)=0\), find the constant for \(\displaystyle y_1(t)\). $$ 0 = y(0) = y_1(0) = e^{-2(0)}\left(\int e^{2(0)}\,dt + C\right) = 1 \cdot (0 + C) $$ Thus, \(\displaystyle C=0\). Now, find the constant for \(\displaystyle y_{2}( t)\) by applying the continuity condition at \(\displaystyle t=1\): $$ y_1(1) = y_2(1) $$ Since \(\displaystyle C=0\) for \(\displaystyle y_{1}( t)\): $$ y_1(1) = e^{-2(1)}\left(\int_0^1 e^{2t}\,dt\right) = e^{-2} \left[\frac{1}{2}e^{2}\right] = -\frac{1}{e} $$ Now use this result to find the constant for \(\displaystyle y_{2}( t)\): $$ y_2(1) = e^{-2(1)}(0 + C) = -\frac{1}{e} $$ Thus, \(\displaystyle C=-\frac{1}{e}\). The final piecewise defined solution to the given IVP is: $$ y(t) =\begin{cases} e^{-2t}\left(\int_0^t e^{2\tau}\,d\tau \right) & 0 \leq t \leq 1 \\ e^{-2t}-\frac{1}{e} & t > 1 \end{cases} $$

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations is a problem where we're tasked with finding a function that satisfies a given differential equation and meets specific initial conditions. In simple terms, an initial condition provides a starting point or a specific value that the solution must pass through.
In our case, the IVP is \[y^{\prime}+2y=g(t), \quad y(0)=0\] This tells us that we need to find a function \(y(t)\) such that when \(t = 0\), \(y(0) = 0\). By substituting back into the differential equation, we can verify whether our solution correctly models the given problem statement from the start which is sometimes a challenge due to the nature of differential equations even if it might be straightforward as a rule of thumb.
The role of the initial value helps uniquely determine the solution to a differential equation. Without an initial value, many different solutions can satisfy the differential equation. Hence, the initial value helps in narrowing down to a single curve among many possibilities.

Integrating Factor

The Integrating Factor is a technique used to solve first-order linear ordinary differential equations. The essence of this method is to multiply the differential equation by a carefully chosen function to make the equation easier to solve, often allowing it to be rewritten in a form that can be integrated directly.
For the equation in the exercise, \[y^{\prime}+2y=g(t)\] an integrating factor \( e^{\int 2 dt} = e^{2t} \) is employed. The choice of the integrating factor revolves around the coefficient of \(y\) in the equation (which is 2 in this instance). Once this factor multiplies the entire equation, it turns the left-hand side into the derivative of a product of functions, thus simplifying integration.
Post multiplication by the integrating factor, both sides of the equation can be integrated easily. The equation transforms into something involving the derivative of \(y\) times the integrating factor, which can essentially be simplified into: \[ y e^{2t} = \int e^{2t} g(t) \, dt + C \]
The solution eventually reveals the function \(y(t)\), showing how integrating factors are a clever technological move in differential equation solving that sidesteps more complex algebra.

Piecewise Function

A piecewise function is a function defined by multiple distinct equations, each applying to a certain interval of the main input variable. The function value and behavior can change at specific points, making this an excellent way of modeling scenarios which have different conditions at various segments.
In this problem, the function \(g(t)\) is piecewise defined: \[g(t)=\begin{cases}1, & 0 \leq t \leq 1 \0, & t>1\end{cases}\] This means that for \(t\) between 0 and 1, \(g(t)\) takes a value of 1; beyond that timeframe, it switches to a value of 0. Piecewise functions are important in mathematical modeling because they can effectively describe situations where a system behaves in different ways under different conditions or stimuli.
Within our solution steps, these distinct intervals necessitate separate solutions, namely \(y_1(t)\) and \(y_2(t)\), reflecting how \(y(t)\) responds to \(g(t)\) at different intervals. When solving differential equations, piecewise behaviors of functions often lead to creating and combining solutions from these distinct parts to form a complete picture.

Continuity Condition

Continuity Condition is an important concept when dealing with piecewise functions and solutions to differential equations. It ensures that there is seamless connection between various parts of the function, especially when solutions for those parts are obtained separately.
For our solution, continuity condition comes into play at \(t=1\). At this point, we expect the solutions \(y_1(t)\) and \(y_2(t)\) to converge, ensuring the entire function remains continuous across its interval, i.e., \[y_1(1) = y_2(1)\]
By managing continuity, the overall behavior of the function is smooth as one segment transitions into the next. Applying this condition allowed for determining the constant in \(y_2(t)\) using the known value at the endpoint of its previous segment, \(y_1(1)\).
Continuity conditions not just maintain a smooth graph, but they also ensure that the mathematical model accurately reflects real-world phenomena in a coherent manner, without improbable jumps or breaks in function outputs upon evaluation.