Question:

Let \(y=y(t)\) be a solution of the differential equation\(\frac{d y}{d t}+\alpha y=\gamma e^{-\beta t}\) where, \(\alpha > 0, \beta>0\) and \(\gamma > 0\). Then \(\displaystyle\lim _{t \rightarrow \infty} y(t)\)

Updated On: Jan 8, 2025
  • is$-1$
  • is 0
  • is 1
  • does not exist
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The Correct Option is B

Approach Solution - 1

The given differential equation is:

\( \frac{dy}{dt} + \alpha y = y e^{-\beta t} \),

where \(\alpha\), \(\beta\), \(\gamma\) are constants. We are given that:

\( \lim_{t \to \infty} y(t) = 0. \)

Step 1: Rewrite the equation

Factoring \(y\) from the right-hand side:

\( \frac{dy}{dt} = y(e^{-\beta t} - \alpha). \)

Dividing through by \(y\) (assuming \(y \ne 0\)):

\( \frac{1}{y} \frac{dy}{dt} = e^{-\beta t} - \alpha. \)

Step 2: Solve the differential equation

Rewriting:

\( \frac{dy}{y} = (e^{-\beta t} - \alpha) dt. \)

Integrating both sides:

The left-hand side integrates to:

\( \int \frac{1}{y} dy = \int (e^{-\beta t} - \alpha) dt. \)

\( \ln |y| = \int e^{-\beta t} dt - \alpha t + C, \)

where \(C\) is the constant of integration.

For the first term on the right-hand side:

\( \int e^{-\beta t} dt = -\frac{1}{\beta} e^{-\beta t}. \)

Thus:

\( \ln |y| = -\frac{1}{\beta} e^{-\beta t} - \alpha t + C. \)

Exponentiating both sides:

\( y = e^{-\frac{1}{\beta} e^{-\beta t} - \alpha t + C}. \)

Simplify the exponent:

\( y = e^C \cdot e^{-\frac{1}{\beta} e^{-\beta t}} \cdot e^{-\alpha t}. \)

Let \(e^C = K\) (a constant). Then:

\( y = K \cdot e^{-\frac{1}{\beta} e^{-\beta t}} \cdot e^{-\alpha t}. \)

Step 3: Analyze \(y(t)\) as \(t \to \infty\)

1. As \(t \to \infty\), \(e^{-\beta t} \to 0\). Therefore, the term \(e^{-\frac{1}{\beta} e^{-\beta t}} \to e^0 = 1\). 2. The dominant term in \(y(t)\) is \(e^{-\alpha t}\), which goes to 0 as \(t \to \infty\), provided \(\alpha > 0\).

Thus:

\( \lim_{t \to \infty} y(t) = 0. \)

Final Answer: The solution satisfies the condition \(\lim_{t\rightarrow\infty} y(t) = 0\), and the correct option is: 2 (0).

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Approach Solution -2

The correct answer is (B) : is 0


Solution


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Concepts Used:

Types of Differential Equations

There are various types of Differential Equation, such as:

Ordinary Differential Equations:

Ordinary Differential Equations is an equation that indicates the relation of having one independent variable x, and one dependent variable y, along with some of its other derivatives.

\(F(\frac{dy}{dt},y,t) = 0\)

Partial Differential Equations:

A partial differential equation is a type, in which the equation carries many unknown variables with their partial derivatives.

Partial Differential Equation

Linear Differential Equations:

It is the linear polynomial equation in which derivatives of different variables exist. Linear Partial Differential Equation derivatives are partial and function is dependent on the variable.

Linear Differential Equation

Homogeneous Differential Equations:

When the degree of f(x,y) and g(x,y) is the same, it is known to be a homogeneous differential equation.

\(\frac{dy}{dx} = \frac{a_1x + b_1y + c_1}{a_2x + b_2y + c_2}\)

Read More: Differential Equations