Basic First ODE and Separation problem

In summary: I agree that the homogeneous version ##y' + y = 0## is separable, but perhaps even more relevantly, it has an integrating factor ##e^x## such that ##d(e^x y)/dx = e^x(y' + y)##. Multiplying both sides by ##e^x## therefore gives ##d(e^x y)/dx = e^{5x}##, which is easily solvable by integrating both sides.
  • #1
Gulfstream757
3
0
<Moderator's note: Moved from a technical forum and thus no template.>

Hi all, hope you all had a great Christmas,

I had a difficulty with theses two problems,

1. Show that y=1/x + x/2 is a solution of the differential equation,

dy/dx=1-y/x, where y=1.5 and x=2
2. Solve the differential equation, dy/dx=e^4x-y, where y=1 and x=1/4
In your final answer express y in terms of x.


For question 1 my understanding is that I should differentiate the y= function and then sub that in for dy/dx and they should equal with some algebra. My problem is when do you input the values for x and y?

For question 2 I think its a separable equation that you would separate, integrate to solve then sub in the x and y values to find the constant of integration. My issue is how to separate, I've tried to take the natural log of both sides? But I am really lost

Thanks so much in advance!
 
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  • #2
Gulfstream757 said:
where y=1.5 and x=2
This is pretty bad nomenclature because y and x are not constants so it is not that you should insert this into the differential equation. Instead, it would seem that this is supposed to give the boundary condition, i.e., that y = 1.5 when x = 2. You then need to show that the function y both satisfies the differential equation as well as the boundary condition (a priori, you could have been given a function that satisfies the differential equation but not the boundary condition). Hence, you need to check that both the differential equation and the boundary condition are satisfied.

Gulfstream757 said:
For question 2 I think its a separable equation that you would separate, integrate to solve then sub in the x and y values to find the constant of integration. My issue is how to separate, I've tried to take the natural log of both sides? But I am really lost
It is not clear to me what you mean by e^4x-y. Do you mean e^(4x)-y, (e^4)x - y, e^(4x-y), or something else? Please use parentheses or LaTeX to avoid any misunderstanding. Also, please show what you tried and what you got from trying.
 
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  • #3
Gulfstream757 said:
<Moderator's note: Moved from a technical forum and thus no template.>

Hi all, hope you all had a great Christmas,

I had a difficulty with theses two problems,

1. Show that y=1/x + x/2 is a solution of the differential equation,

dy/dx=1-y/x, where y=1.5 and x=2
2. Solve the differential equation, dy/dx=e^4x-y, where y=1 and x=1/4
In your final answer express y in terms of x.


For question 1 my understanding is that I should differentiate the y= function and then sub that in for dy/dx and they should equal with some algebra. My problem is when do you input the values for x and y?

For question 2 I think its a separable equation that you would separate, integrate to solve then sub in the x and y values to find the constant of integration. My issue is how to separate, I've tried to take the natural log of both sides? But I am really lost

Thanks so much in advance!

If your second equation is ##dy/dx = e^{4x} - y## then it is, indeed, "separable", and re-writing it as ##dy/dx + y = e^{4x}## makes it clearer. If your second equation is ##dy/dx = e^{4x-y}## then it is not separable in any standard sense.

The way you wrote it, the first form is what one would get applying standard math parsing rules; however, using parentheses never hurts and can sometimes help.
 
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  • #4
Ray Vickson said:
If your second equation is ##dy/dx = e^{4x} - y## then it is, indeed, "separable", and re-writing it as ##dy/dx + y = e^{4x}## makes it clearer.
That is not a separable ODE. A separable ODE can be written on the form ##y' f(y) = g(x)##, where ##f## and ##g## are some functions. You can then solve the differential equation by integrating from ##x_0## to ##x## according to
$$
\int_{x_0}^x g(\xi) d\xi = \int_{x_0}^x y'(\xi) f(y(\xi)) d\xi = \int_{y(x_0)}^{y(x)} f(y) dy,
$$
where the last step is a change of variables in the integral.

If your second equation is ##dy/dx = e^{4x-y}## then it is not separable in any standard sense.

That, on the other hand, is separable. I will not do it as it is something the OP should do for himself.
 
  • #5
Gulfstream757 said:
2. Solve the differential equation, dy/dx=e^4x-y
The expression on the right side of the equation is ambiguous. Because many new members here don't realize the importance of using parentheses, it's hard for us to know exactly what you mean when you write e^4x-y.

This could be any of the following:
##e^4~x - y##
##e^{4x} - y##
##e^{4x - y}##
I've written these using LaTeX. Writing the 2nd and 3rd as inline text, they would be e^(4x) - 4 and e^(4x - y).
 
  • #6
Orodruin said:
That is not a separable ODE. A separable ODE can be written on the form ##y' f(y) = g(x)##, where ##f## and ##g## are some functions. You can then solve the differential equation by integrating from ##x_0## to ##x## according to
$$
\int_{x_0}^x g(\xi) d\xi = \int_{x_0}^x y'(\xi) f(y(\xi)) d\xi = \int_{y(x_0)}^{y(x)} f(y) dy,
$$
where the last step is a change of variables in the integral.
That, on the other hand, is separable. I will not do it as it is something the OP should do for himself.

You are right: I am somewhat sloppily using the term "separable" when looking at a nonhomogeneous DE whose homegeneous version is separable, although, of course, it is not separable officially. And, of course, I spoke too soon about the second DE. Blame the lack of morning coffee.
 
  • #7
Ray Vickson said:
You are right: I am somewhat sloppily using the term "separable" when looking at a nonhomogeneous DE whose homegeneous version is separable, although, of course, it is not separable officially.
I agree that the homogeneous version ##y' + y = 0## is separable, but perhaps even more relevantly, it has an integrating factor ##e^x## such that ##d(e^x y)/dx = e^x(y' + y)##. Multiplying both sides by ##e^x## therefore gives ##d(e^x y)/dx = e^{5x}##, which is easily integrated.
 
  • #8
Orodruin said:
This is pretty bad nomenclature because y and x are not constants so it is not that you should insert this into the differential equation. Instead, it would seem that this is supposed to give the boundary condition, i.e., that y = 1.5 when x = 2. You then need to show that the function y both satisfies the differential equation as well as the boundary condition (a priori, you could have been given a function that satisfies the differential equation but not the boundary condition). Hence, you need to check that both the differential equation and the boundary condition are satisfied.It is not clear to me what you mean by e^4x-y. Do you mean e^(4x)-y, (e^4)x - y, e^(4x-y), or something else? Please use parentheses or LaTeX to avoid any misunderstanding. Also, please show what you tried and what you got from trying.
Sorry for only getting a chance to reply now, I've attached my attempts, I know it looks like I haven't put i
Orodruin said:
This is pretty bad nomenclature because y and x are not constants so it is not that you should insert this into the differential equation. Instead, it would seem that this is supposed to give the boundary condition, i.e., that y = 1.5 when x = 2. You then need to show that the function y both satisfies the differential equation as well as the boundary condition (a priori, you could have been given a function that satisfies the differential equation but not the boundary condition). Hence, you need to check that both the differential equation and the boundary condition are satisfied.It is not clear to me what you mean by e^4x-y. Do you mean e^(4x)-y, (e^4)x - y, e^(4x-y), or something else? Please use parentheses or LaTeX to avoid any misunderstanding. Also, please show what you tried and what you got from trying.
Apologies I only got a chance to reply now, please see the photos attached, as you can see I'm really stuck! Thanks in advance for any help!

26241101_629806707410224_980919250_n.jpg


26241515_629806730743555_221343950_n.jpg
 

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  • #9
The equation as written is separable, so you should start by separating it. Once you have done that it canbe integrated without problems in the manner described in post #4.
 
  • #10
Gulfstream757 said:
Sorry for only getting a chance to reply now, I've attached my attempts, I know it looks like I haven't put i
Apologies I only got a chance to reply now, please see the photos attached, as you can see I'm really stuck! Thanks in advance for any help!

View attachment 217799

View attachment 217800

You are already getting into a bad habit: posting images instead of typed work. Please avoid that in future, especially if you are serious about wanting help. (See the post "guidelines for students and helpers", by Vela, which is pinned to the start of this forum, for various reasons related to this issue.)
 

1. What is a basic first order differential equation (ODE)?

A basic first order ODE is a mathematical equation that relates an unknown function and its derivative. It is typically in the form of dy/dx = f(x), where y is the unknown function and f(x) is a function of x.

2. What is the separation problem in ODEs?

The separation problem in ODEs refers to the process of solving an ODE by separating the variables on opposite sides of the equation. This involves isolating the dependent variable on one side and the independent variable on the other side, and then integrating both sides to find a general solution.

3. How do you solve a basic first order ODE using separation of variables?

To solve a basic first order ODE using separation of variables, first separate the variables on opposite sides of the equation. Then, integrate both sides to find a general solution. Finally, use initial conditions to determine the particular solution.

4. What are initial conditions in ODEs?

Initial conditions refer to the conditions or values of the unknown function and its derivative at a specific point. These conditions are used to determine the particular solution of an ODE, as the general solution may have multiple possible solutions.

5. What are some real-world applications of basic first order ODEs?

Basic first order ODEs have many real-world applications, such as in physics, engineering, economics, and biology. They can be used to model various physical systems, such as population growth, radioactive decay, and fluid dynamics. They also play a crucial role in solving optimization problems and predicting future behavior in different fields.

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