Problem 35 (a) What value \(y=\) constant s... [FREE SOLUTION]

Chapter 6: Problem 35

(a) What value \(y=\) constant solves \(d y / d t=-2 y+12\) ? (b) Find the solution with an arbitrary constant \(A\). (c) What solutions start from \(y_{0}=0\) and \(y_{0}=10 ?\) (d) What is the steady state \(y_{\infty}\) ?

Short Answer

Expert verified

a) y = 6 b) y = 6 + Ae^{-2t} c) y = 6 - 6e^{-2t}; y = 6 + 4e^{-2t} d) y_{\infty} = 6

Step by step solution

Solve the Constant Solution for y

For part (a), we're looking for a constant solution where \(dy/dt = 0\). Substituting into the equation gives \(-2y + 12 = 0\). Solving for \(y\), we find \(y = 6\).

Solve the Differential Equation Generically

For part (b), the differential equation is \(dy/dt = -2y + 12\). We can transform it into a separable equation: \(dy/(12-2y) = dt\). Integrating both sides, we get \(-\frac{1}{2} \ln|12-2y| = t + C\), where \(C\) is the integration constant. Solving for \(y\), we have the solution \(y = 6 + Ae^{-2t}\), with \(A\) as an arbitrary constant determined by initial conditions.

Apply Initial Conditions

For part (c), using the general solution \(y = 6 + Ae^{-2t}\), apply the initial conditions to find specific solutions. - If \(y_0 = 0\), then \(y = 6 - 6e^{-2t}\).- If \(y_0 = 10\), solve \(10 = 6 + A\) to yield \(A = 4\). Thus, \(y = 6 + 4e^{-2t}\).

Determine Steady State Solution

For part (d), the steady state \(y_{\infty}\) is the value \(y\) approaches as \(t\) goes to infinity. From the solution \(y = 6 + Ae^{-2t}\), as \(t \to \infty\), \(e^{-2t} \to 0\), leading to \(y \to 6\). Therefore, the steady state is \(y_{\infty} = 6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Solutions

In differential equations, a constant solution refers to a solution where the rate of change, or derivative, of the variable is zero. This means the function does not change with time. In our exercise, we are asked to find a constant value for the differential equation \(dy/dt = -2y + 12\).
To find this, we set the derivative equal to zero, \(dy/dt = 0\). This simplifies the equation to \(-2y + 12 = 0\). Solving for \(y\), we find that the constant solution is \(y = 6\). This means if \(y = 6\), the system remains unchanged over time.
Constant solutions often represent equilibrium points or states where a system is balanced.

Initial Conditions

Initial conditions are values given at the start of a problem which allow us to determine a unique solution for a differential equation that contains arbitrary constants.
Using the general solution derived from our separable differential equation, \(y = 6 + Ae^{-2t}\), we can apply different initial conditions to find particular solutions.

If \(y_0 = 0\) at \(t = 0\), then setting \(y = 0\) gives \(0 = 6 + A\), or \(A = -6\). Therefore, the specific solution is \(y = 6 - 6e^{-2t}\).
For \(y_0 = 10\), applying the initial condition yields \(10 = 6 + A\), or \(A = 4\). Thus, the specific solution becomes \(y = 6 + 4e^{-2t}\).

Initial conditions are crucial in mathematical modeling as they set the starting point for analyzing the system鈥檚 behavior over time.

Steady State

The steady state of a differential equation refers to a situation where the variable reaches a state of equilibrium such that its derivative becomes zero over time. This means the value doesn't change as time progresses further.
To determine the steady state value \(y_\infty\) for our problem, we examine the behavior of the solution \(y = 6 + Ae^{-2t}\) as time \(t\) approaches infinity.

The term \(e^{-2t}\) approaches zero as \(t\) becomes very large because the exponential function decays to zero, making \(y\) approach 6.
Thus, the steady state of the solution is \(y_\infty = 6\).

Identifying the steady state is important as it helps in understanding the long-term behavior of dynamic systems.

Separable Differential Equations

Separable differential equations are a type of differential equation that can be rearranged and expressed in a form where the variables can be separated on opposite sides of the equation. This makes them easier to solve using integration.
In the provided exercise, we have the equation \(dy/dt = -2y + 12\). To solve it, it's rewritten as \(dy/(12-2y) = dt\), isolating terms involving \(y\) on one side and \(t\) on the other. Once separated, each side can be integrated individually:

Integrate \(dy/(12-2y)\) to find \(-\frac{1}{2} \ln|12-2y|\).
Integrate \(dt\) to find \(t + C\), where \(C\) is the integration constant.

This leads to the general solution \(y = 6 + Ae^{-2t}\). Solving separable differential equations this way allows us to find a wide range of potential solutions subject to different conditions.

91影视

(a) What value \(y=\) constant solves \(d y / d t=-2 y+12\) ? (b) Find the solution with an arbitrary constant \(A\). (c) What solutions start from \(y_{0}=0\) and \(y_{0}=10 ?\) (d) What is the steady state \(y_{\infty}\) ?

Short Answer

Step by step solution

Solve the Constant Solution for y

Solve the Differential Equation Generically

Apply Initial Conditions

Determine Steady State Solution

Key Concepts

Constant Solutions

Initial Conditions

Steady State

Separable Differential Equations

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