Differential equation problems

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Homework Help Overview

The discussion revolves around two differential equation problems. The first involves a separable equation with initial conditions, while the second is a first-order linear differential equation. Participants are exploring methods to approach these problems and clarify their understanding of the concepts involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to separate variables in the first differential equation but expresses uncertainty about the algebra involved. They also seek clarification on the form of the second equation and the role of p(x). Some participants question the classification of the first equation as separable and suggest alternative approaches.

Discussion Status

Participants are actively engaging with the problems, offering hints and algebraic manipulations for the first equation. There is a mix of interpretations regarding the separability of the first equation, and some guidance has been provided on the second equation's structure. However, no consensus has been reached on the best approach to either problem.

Contextual Notes

There is a discussion about the definitions and classifications of the differential equations, particularly regarding the separability and the forms of the equations. Participants are also navigating the complexities of integration techniques required for the second equation.

antiflag403
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Differential equation problems...

Hey everyone,
Having some problems with these differential equations. If someone could point me in the right direction I would be really grateful.
1) dy/dx=(5x+12y)/3x with initial conditions y(1)=5.
I know this is a seperable equation but I'm not sure of the correct algebra to separate them. Do I make it 5x/3x +12y/3x and then work from there??
2) dx/dt+3x=cos(2t) x(0)=-1
I know this one is a first order linear differential equation so it must be in the form dx/dt+p(x)y=q(x) and then you take e^(integral)p(x) and so on. I'm not sure what to do since there is no p(x) (or is it just 3? integral being 3x). I tried that but I have trouble with the integration by parts that results.
If anyone could give me some advice or point me in the right direction that would be great.
Thanks
 
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antiflag403 said:
Hey everyone,
Having some problems with these differential equations. If someone could point me in the right direction I would be really grateful.
1) dy/dx=(5x+12y)/3x with initial conditions y(1)=5.
I know this is a seperable equation but I'm not sure of the correct algebra to separate them. Do I make it 5x/3x +12y/3x and then work from there??
Then you do not know what a "separable" equation is! How would you propose to "separate" x and y?

2) dx/dt+3x=cos(2t) x(0)=-1
I know this one is a first order linear differential equation so it must be in the form dx/dt+p(x)y=q(x) and then you take e^(integral)p(x) and so on. I'm not sure what to do since there is no p(x) (or is it just 3? integral being 3x). I tried that but I have trouble with the integration by parts that results.
If anyone could give me some advice or point me in the right direction that would be great.
Thanks
No, it's not of the form dx/dt+ p(x)y= q(x)! It is of the form
dx/dt+ p(t)x= q(t). And then, Yes, p(t)= 3.
\int e^{3t}cos(2t)dt requires integration by parts twice and then think!
 
well if the first one is not seperable then I don't know of a way to solve it. Could you give me a hint. Thanks for the help by the way.
 
Here's some help for the 1st differential equation...

Some algebraic manipulation will give you the expression

\frac{dy}{dx} - \frac{4}{x} y = \frac{5}{3}

Do you know how to solve it now?

All the best!
 
The first is, in fact, separable, if you make a change of variables:
\frac{dy}{dx}=\frac{5x+12y}{3x}=\frac{5}{3}+4\frac{y}{x}=f(\frac{y}{x})

Introduce the variable:
u=\frac{y}{x}\to\frac{du}{dx}=\frac{\frac{dy}{dx}-u}{x}\to\frac{1}{(f(u)-u)}\frac{du}{dx}=\frac{1}{x}

In this case, of course, we have: \frac{dy}{dx}=f(u)=\frac{5}{3}+4u
 
Last edited:

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