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Chapter 5: Problem 10

Determine the growth constant \(k\), then find all solutions of the given differential equation. $$5 y^{\prime}-6 y=0$$

Short Answer

Expert verified
k = \frac{6}{5}, y(t) = Ce^{\frac{6}{5} t}

Step by step solution

01

Rewrite the Differential Equation

First, rewrite the differential equation in standard form. We want to express it as \[ \frac{dy}{dt} + P(t) y = 0 \]Given equation: \[ 5y' - 6y = 0 \]Divide through by 5 to get it to standard form: \[ y' - \frac{6}{5} y = 0 \]
02

Identify Growth Constant

The given differential equation now is:\[ y' - \frac{6}{5} y = 0 \]Here, the coefficient of y is the growth constant.Therefore, \[ k = \frac{6}{5} \]
03

Solve the Differential Equation

The general solution for a first-order linear differential equation of the form\[ y' - ky = 0 \]is:\[ y(t) = Ce^{kt} \]Here, \( k = \frac{6}{5} \). Therefore, the solution is:\[ y(t) = Ce^{\frac{6}{5} t} \]

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Key Concepts

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

first-order linear differential equation
A first-order linear differential equation is a type of differential equation where the highest derivative is the first derivative. The general form is: \( \frac{dy}{dt} + P(t)y = Q(t) \)
This equation is linear because neither the function \(y\) nor its first derivative \( \frac{dy}{dt} \) is raised to a power higher than one.
In the exercise, the given equation is \(5y' - 6y = 0\).
To put it in the standard form, we divide by 5.
Now it looks like:\( y' - \frac{6}{5}y = 0\).

The left side of this equation shows the typical linear structure with \(y'\) and \(y\). Notice that there are no products or powers of \(y\) and \(y'\), making it linear.
growth constant
The growth constant is a key parameter in differential equations related to growth and decay processes. It determines the rate at which solutions grow or decay.
Once the differential equation is in standard form, the growth constant is the coefficient of \(y\).
For the equation given in the exercise:\( y' - \frac{6}{5} y = 0 \)
The growth constant, \(k\), is \(\frac{6}{5} \).
This means that the rate of growth is proportional to \(\frac{6}{5} = 1.2\), or a 120% increase over time.
Understanding \(k\) gives insight into the nature of the solution: the larger \(k\), the faster the growth.
general solution
The general solution to a first-order linear differential equation is found using an integrating factor or solving through separation of variables. For a standard form \( y' - ky = 0 \), the solution is:
\( y(t) = Ce^{kt} \).
Here, \(C\) is an arbitrary constant determined by initial conditions.
In the exercise, we already identified \( k = \frac{6}{5} \).
The general solution is:
\( y(t) = Ce^{\frac{6}{5} t} \).
This exponential solution describes how \(y\) grows over time with rate \( k \left( \frac{6}{5} \right)\).Utilize initial conditions, if available, to solve for the constant \(C\).

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Most popular questions from this chapter

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