Solving Integral and ODE Problem

[mmh37]

1. Problem with Integral

I got stuck on the following Integal and can't find the mistake:

$$Integral(x/(x-1))$$

let U' = x and V = (x-1)^(-1)
and U= 0.5x^2 V' = 1/(x-1)^2

so: $$x^2/(2(x-1))-1/2Integral(x^2/(x-1)^2$$

algebraic division: $$x^2 : (x^2 -2x + 1) = 1- 1/(x-1)^2 + 1/(x-1)^2$$

substituting back into equation gives:
$$1/2 [x^2/(x-1) - x - ln (x-1)^2 x 1/(x-1)] = (x+1)/(2(x-1)) + ln(x-1)$$

but the result is supposed to be:
$$x + ln(x-1)$$

2. Substitution in ODE

$$xydy/dx + (x^2 + y^2 +x) = 0$$

I have tried to substitute z = xy and also z = x^2 + y^2, but that didn't work. Any other ideas?

Thank you!

 

[mmh37]

PS: How can I write integrals with the [tex] function?

 

[Tx]

For the first, Plus and Minus one from the top. This will give:
I (x - 1 + 1)/ (x - 1) dx
I 1dx + I 1/(x-1) dx
x + ln(x-1) + C
#

The second, I got:
y^2 = x(1-x)/(1+x) + C
However, I am new to ODE's so it's probably wrong.

 

[mmh37]

*lol* thanks... that was easy... I seem to have a tendency to make things more complicated than they actually are...

But thanks again!

 

[dicerandom]

BTW, to do integrals in TeX:

\int f(x) dx
$$\int f(x) dx$$

\int_0^5 f(x) dx
$$\int_0^5 f(x) dx$$

 

[saltydog]

Hey mmy, the second one IS complicated. Just for the record in case you haven't figured it out, make it exact by finding an integrating factor. You know partial this, partial that, some arithmetic, get x. Solving, I get:

$$y(x)=\pm\frac{\sqrt{\frac{k}{x^2}-3x^2-4x}}{\sqrt{6}}$$

with the sign dictated by the initial conditions.

 

[twoflower]

I got

$$y(x) = \frac{1}{x}\sqrt{C - \frac{1}{2}x^4 - \frac{2}{3}x^3}$$

 

[twoflower]

I got
$$y(x) = \frac{1}{x}\sqrt{C - \frac{1}{2}x^4 - \frac{2}{3}x^3}$$

Which is the same as saltydog has, I see :)