Question

Find the value of \(y_{0}\) for which the solution of the initial value problem $$ y^{\prime}-y=1+3 \sin t, \quad y(0)=y_{0} $$ remains finite as \(t \rightarrow \infty\).

Short answer

Answer: The value of \(y_0\) is \(-4\).

Step-by-step solution

Solve the differential equation using integrating factor

First, rewrite the given differential equation: $$ y^{\prime}-y=1+3 \sin t $$ This is a first-order linear ordinary differential equation in the form of: $$ y^{\prime} + P(t)y = Q(t) $$ Here, we have \(P(t)=-1\) and \(Q(t)=1+3\sin t\). Now, we will find the integrating factor (IF) which is given by \(e^{\int P(t) \, dt}\). Calculate the integrating factor: $$ IF = e^{\int -1 dt} = e^{-t} $$ Now, multiply both sides of the differential equation with the integrating factor: $$ e^{-t}y^{\prime} - e^{-t}y = e^{-t}(1+3\sin t) $$ Now, observe that the left-hand side of the equation is the derivative of the product \(y(t)e^{-t}\): $$ \frac{d}{dt}(y(t)e^{-t}) = e^{-t}(1+3\sin t) $$ Now, we integrate both sides with respect to t: $$ \int \frac{d}{dt}(y(t)e^{-t}) dt = \int e^{-t}(1+3\sin t) dt $$

Find the function \(y(t)\) and its limit as \(t \rightarrow \infty\)

Using integration by parts (twice) on the right-hand side of the equation and remembering that the integral of the derivative of a function is the function itself, we have: $$ y(t)e^{-t} = -e^{-t} - 3e^{-t}\cos t + C $$ To find the general solution, multiply both sides by \(e^t\) to clear out the exponential term: $$ y(t) = -1 - 3\cos t + Ce^t $$ The problem asks for the solution to remain finite as \(t \rightarrow \infty\). To satisfy this condition, we need \(Ce^t\) to go to zero as \(t \rightarrow \infty\). Therefore, we must have \(C=0\). Thus, we get the specific solution: $$ y(t) = -1 - 3\cos t $$

Use the initial condition to find the value of \(y_0\)

Use the initial condition \(y(0) = y_0\): $$ y_0 = -1 - 3\cos (0) $$ Since \(\cos (0) = 1\), we get: $$ y_0 = -1 - 3 \cdot 1 = -4 $$ So the value of \(y_0\) for which the solution remains finite as \(t \rightarrow \infty\) is \(y_0 = -4\).

Key concepts

First-Order Linear Ordinary Differential Equation

First-order linear ordinary differential equations are a specific type of differential equation. They have the general form:

• \( y' + P(t)y = Q(t) \)

Where:

• \( y' \) is the derivative of \( y \) with respect to \( t \).
• \( P(t) \) and \( Q(t) \) are given functions of \( t \). They can be constant or vary with \( t \).

In our exercise, we rewrite the equation as:

• \( y' - y = 1 + 3 \sin t \)

In this form, we can see that it matches \( y' + P(t)y = Q(t) \). Here, \( P(t) = -1 \) and \( Q(t) = 1 + 3 \sin t \). To solve this, we need to employ techniques such as finding an integrating factor. This transforms the equation into something simpler to work with.

Integrating Factor

The integrating factor is a key tool in solving first-order linear differential equations. Think of it as a magic multiplier that helps simplify the equation. To find it, use the formula:

• \( IF = e^{\int P(t) \ dt} \)

This integrating factor makes the product \( y(t)e^{-t} \) easier to handle. In our problem, the integrating factor is computed as:

• \( IF = e^{-t} \)

Once you multiply every term in the differential equation by this integrating factor, the left side of the equation transforms into the derivative of a product. It simplifies to:

• \( \frac{d}{dt}(y(t)e^{-t}) = e^{-t}(1+3\sin t) \)

This is much easier to integrate on both sides, ultimately leading us towards solving for \( y(t) \).

Integration by Parts

Integration by parts is a technique used when integrating the product of two functions. In this exercise, we employed integration by parts twice. This was necessary because the expression \( e^{-t} (1+3\sin t) \) required it for a complete solution. The integration by parts formula is:

• \( \int u \, dv = uv - \int v \, du \)

In our solution:

1. First integration yields a simpler form after applying the formula.
2. The second round of integration by parts further simplifies the expression.

Eventually, we arrive at:

• \( y(t)e^{-t} = -e^{-t} - 3e^{-t} \cos t + C \)

This expression, once simplified using integration by parts, allows us to isolate \( y(t) \).

Bounded Solutions

When looking at differential equations, especially initial value problems, one important aspect is whether solutions remain bounded. A bounded solution is one that does not go to infinity as time \( t \) approaches infinity.
This ensures predictability and stability of the solution over time.In our problem, when \( y(t) = -1 - 3\cos t + Ce^t \), for the solution to be bounded, \( Ce^t \) must approach zero as \( t \to \infty \).

• Therefore, \( C \) must be zero.
• If \( C eq 0 \), \( Ce^t \) becomes unbounded because \( e^t \) grows exponentially.

Finally, solving \( y(0) = y_0 \) gives \( y_0 = -4 \), ensuring the solution remains finite.
This shows how elements such as initial conditions affect the nature of solutions.