Reid's question at Yahoo Answers regarding a first order separable IVP

In summary, a first order separable IVP is a type of ordinary differential equation that can be separated into two parts and includes an initial condition. To solve it, the equation is separated into two parts and integrated, with the initial condition used to solve for the unknown constant. These types of equations are important in science and engineering as they can model and predict real-world phenomena. Some common applications include population growth, chemical reactions, and electrical circuits. However, first order separable IVPs have some limitations, such as only being able to solve linear equations with first degree derivatives and potentially not accurately predicting behavior in complex systems.
  • #1
MarkFL
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Here is the question:

What is the particular solution to y'x-1=0 using the inition condition x=1,y=2?

I have posted a link there to this topic so the OP can see my work.
 
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  • #2
Hello reid,

We are given the IVP:

\(\displaystyle y'x-1=0\) where \(\displaystyle y(1)=2\)

Separating the variables, we obtain:

\(\displaystyle dy=\frac{1}{x}\,dx\) where \(\displaystyle x\ne0\)

Switching the dummy variables of integration and utilizing the boundaries as limits, we obtain:

\(\displaystyle \int_2^{y(x)}\,du=\int_1^x\frac{dv}{v}\)

Applying the anti-derivative form of the FTOC, we find:

\(\displaystyle y(x)-2=\ln|x|\)

\(\displaystyle y(x)=\ln|x|+2\)
 

FAQ: Reid's question at Yahoo Answers regarding a first order separable IVP

What is a first order separable IVP?

A first order separable IVP (initial value problem) is a type of ordinary differential equation where the highest derivative of the unknown function is of first degree, and it can be separated into two parts: one depending on the function itself, and the other depending on its derivative. It also includes an initial condition, which is the value of the function at a specific point.

How do you solve a first order separable IVP?

To solve a first order separable IVP, you need to separate the equation into two parts, one with the function and the other with its derivative. Then, you can integrate both sides and apply the initial condition to solve for the unknown constant. This will give you the general solution, which can be further simplified if needed.

What is the importance of solving first order separable IVPs?

First order separable IVPs are important in many fields of science and engineering as they represent many real-world phenomena. They allow us to model and predict the behavior of systems that are continuously changing, such as population growth, chemical reactions, and electrical circuits.

What are some applications of first order separable IVPs?

Some common applications of first order separable IVPs include modeling population growth, predicting the concentration of a chemical in a reaction, determining the voltage/current in an electrical circuit, and analyzing the motion of a falling object under the force of gravity.

Are there any limitations to solving first order separable IVPs?

While first order separable IVPs are useful in many situations, they have some limitations. They can only be used to solve linear equations with first degree derivatives, and they may not always provide an accurate prediction of the system's behavior if there are other factors at play. In some cases, more complex methods may be needed to accurately model the system.

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