Solving a Simple Differential Equation with Separation of Variables

[prace]

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.

 

[prace]

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? $$\int{\frac{x^2}{(1+x)}}dx$$

 

[Mute]

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.

 

[prace]

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!

 

[dextercioby]

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.

 

[prace]

Oh cool, even better. Thanks!