Naincy said:Hi. Thanx
No, it is actually 2x and not 2t in the numerator.
It is a function in two variables x and t. I need to integrate it with respect to t. Note that x cannot be considered a constant here.
so , it becomes integration of a function in two variables. i don't know how to start with it.
any help would be appreciated.
Naincy said:Basically, I need to find the solution in x for the following differential equation:
dx/dt = 1 - { 2x/(125 + 3t) }
MarkFL said:Now, we need to compute the integrating factor \(\displaystyle \mu(t)=\exp\left(\int\frac{2}{3t+125}\,dt\right)\)
What do you find?
I am going to move and re-title this thread. :D
Naincy said:Did I get the integrating factor right ?
Naincy said:ok. figured out the mistake. :)
The final answer is:
x(t) = \(\displaystyle \frac{125+3t}{5} + \frac{c}{({125+3t})^{2/3}}\)
I think this is correct. yeah ?
A first order linear ODE (ordinary differential equation) is a mathematical equation that relates a function and its derivatives. It is called "linear" because the function and its derivatives are only raised to the first power and there are no products or powers of the function or its derivatives. The general form of a first order linear ODE is dy/dx + P(x)y = Q(x), where P(x) and Q(x) are functions of x.
The most common method for solving a first order linear ODE is by using an integrating factor. This involves multiplying both sides of the equation by an integrating factor, which is a function of x that makes the left side of the equation equal to the derivative of the product of the integrating factor and the function y. After multiplying and simplifying, the equation can be integrated to solve for y.
An initial value problem is a type of differential equation that has an initial condition, usually written as y(x0) = y0, where x0 is a given value and y0 is the value of the function at that point. In other words, an initial value problem involves finding a particular solution to a differential equation that satisfies a given condition at a specific point.
The main difference between a first order linear ODE and a first order non-linear ODE is the form of the equation. A linear ODE has the form dy/dx + P(x)y = Q(x), while a non-linear ODE has a more complex form, such as dy/dx = P(x)y + Q(x). Non-linear ODEs are generally more difficult to solve and may require numerical methods.
First order linear ODEs are used to model a wide range of physical phenomena, such as population growth, chemical reactions, and electrical circuits. They are also used in engineering and economics to describe relationships between variables. Solving first order linear ODEs allows us to make predictions and understand the behavior of these systems.