Solving a Simple Differential Equation with Separation of Variables

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Homework Help Overview

The problem involves solving a differential equation using the separation of variables method. The original poster presents the equation (x+1)dy/dx = x + 6 and discusses their approach to separating variables and integrating both sides.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to separate variables and integrate, expressing uncertainty about integral techniques. Other participants suggest alternative methods for integration and discuss related integrals, questioning general rules for handling such integrals.

Discussion Status

Participants are actively engaging with the problem, offering insights and alternative approaches to integration. Some guidance has been provided on simplifying integrals, and there is a sense of exploration regarding techniques for similar problems.

Contextual Notes

There is mention of the original poster's uncertainty about basic integral techniques, and participants are exploring different methods without reaching a consensus on a single approach.

prace
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Here is the problem:

Solve, (x+1)\frac{dy}{dx} = x + 6

Here is what I tried:

I moved all the x's to one side and left the dy on the left of the equal sign to solve with the separation of variable method.

I got, \int{dy} = \int{\frac{(x+6)}{(x+1)}dx}

So here I just solve the integrals and I am done. I guess the real question is how do I go about solving the integral on the right? I seem to have forgotten some basic integral techniques.

Thank you.
 
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I came across another integral that I am not catching here that seems to be along the same line as this one. Is there a rule to dealing with these kinds of integrals? \int{\frac{x^2}{(1+x)}}dx
 
For your first integral, you can write \frac{x+6}{x+1} = \frac{(x+1) + 5}{x+1} = 1 + \frac{5}{x+1}which is easy to integrate.

For the second one, you can change variables to u = 1 + x, so du = dx and your integrand becomes


\frac{(u-1)^2}{u} = \frac{u^2 - 2u + 1}{u} = u - 2 + \frac{1}{u},

which again should be easy to integrate.
 
Last edited:
Awesome, that was super easy once you look at it that way. Thanks, now I can look at other integrals and apply the same method. Life somehow just became much easier! ^_^ Thank you!
 
prace said:
I came across another integral that I am not catching here that seems to be along the same line as this one. Is there a rule to dealing with these kinds of integrals? \int{\frac{x^2}{(1+x)}}dx

No need for the substitution hinted.

\frac{x^{2}}{x+1}=x-1+\frac{1}{x+1}

Daniel.
 
Oh cool, even better. Thanks!
 

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