Linear first order differential equations - just a question

In summary, the general solution of the linear first order differential equation is y = \frac{1}{\mu (t)} \left[ \int \mu (s) g(s) \: ds + \mathrm{C} \right], where \mu (t) = \exp \int p(s) \: ds. However, in order to satisfy the initial condition y(t_0) = y_0, the integral limits in the solution need to be adjusted to \int _{t_0} ^t and \int _{t_0} ^{t_0}, respectively. This is because \mu (t_0) = 1 and y(t_
  • #1
DivGradCurl
372
0
I thought the general solution of the linear first order differential equation

[tex]y ^{\prime} + p(t)y = g(t), \qquad y\left( t_0 \right) = y_0[/tex]

were

[tex]y = \frac{1}{\mu (t)} \left[ \int \mu (s) g(s) \: ds + \mathrm{C} \right],[/tex]

where

[tex]\mu (t) = \exp \int p(s) \: ds[/tex]

However, I have found this

[tex]y = \frac{1}{\mu (t)} \left[ \int _{t_0} ^t \mu (s) g(s) \: ds + y_0 \right],[/tex]

where

[tex]\mu (t) = \exp \int _{t_0} ^t p(s) \: ds[/tex]

Can anybody please clarify that? I don't see why those integral limits should be there. Any help is highly appreciated.
 
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  • #2
general solution correct.

however, in this case, integration limits needed for initial condition:

[tex] y\left( t_0 \right) = y_0 [/tex]

that is:

[tex]\mu (t_0) = \exp \int _{t_0} ^{t_0} p(s) \: ds = \exp(0) = 1 [/tex]

[tex]y(t_0) = \frac{1}{\mu (t_0)} \left[ \int _{t_0} ^{t_0} \mu (s) g(s) \: ds + y_0 \right] = \frac {1} {1} \left [(0) \ + \ y_0 \right ] \ = \ y_0 [/tex]

nothing really complicated here
 
  • #3


The general solution of a linear first order differential equation is given by the formula y = \frac{1}{\mu (t)} \left[ \int \mu (s) g(s) \: ds + \mathrm{C} \right], where \mu (t) = \exp \int p(s) \: ds. This formula is correct and can be used to find the general solution to any linear first order differential equation. However, in the specific case where we have an initial condition y(t_0) = y_0, the value of \mathrm{C} can be determined using this condition. This results in the second formula y = \frac{1}{\mu (t)} \left[ \int _{t_0} ^t \mu (s) g(s) \: ds + y_0 \right], where \mu (t) = \exp \int _{t_0} ^t p(s) \: ds. The integral limits are necessary in this case because we are integrating from the initial time t_0 to the current time t. The presence of the initial condition y(t_0) = y_0 also affects the value of \mathrm{C}, hence the difference in the formula. Both formulas are correct and can be used depending on the given initial condition. I hope this clarifies the confusion.
 

1. What is a linear first order differential equation?

A linear first order differential equation is an equation that involves the first derivative (or rate of change) of a function, with the function itself appearing in a linear form. It can be written in the form: y' = f(x,y).

2. How do you solve a linear first order differential equation?

To solve a linear first order differential equation, you can use the method of separation of variables. This involves isolating the variables on opposite sides of the equation and then integrating both sides. You can also use the integrating factor method or the method of undetermined coefficients.

3. What is the difference between linear and nonlinear first order differential equations?

The main difference between linear and nonlinear first order differential equations is that a linear equation has a linear relationship between the function and its derivatives, while a nonlinear equation does not. This means that the derivatives in a linear equation only appear to the first power, whereas in a nonlinear equation, they can appear to any power.

4. What are some real-world applications of linear first order differential equations?

Linear first order differential equations are used in many fields of science and engineering, including physics, chemistry, and economics. They can be used to model a wide range of processes, such as population growth, radioactive decay, and electrical circuits.

5. Can a linear first order differential equation have multiple solutions?

Yes, a linear first order differential equation can have multiple solutions. This is because the general solution to a linear equation will contain a constant of integration, which can take on different values. This means that there can be infinitely many solutions to a linear equation, depending on the initial conditions given.

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