1. ## Homogeneous First-order ODE.

i got one more task but i dont want to start new thread, i solved it but im not quite sure if it's ok so ill put the solution here and if one can take time to review it ill be thankful, so here it is:

the problem is: $\displaystyle xy'-y=sqrt(x^2+y^2)$

i started this way: first get the y on the right side => $\displaystyle xy'=sqrt(x^2+y^2)+y$ now i divide both sides by x so => $\displaystyle y' = (sqrt(x^2+y^2)+y)/x$ now i separate this in two ratios => $\displaystyle y' = sqrt(x^2+y^2)/x + y/x$ next i put the x inside the sqrt so => $\displaystyle y' = sqrt((x^2+y^2)/x^2) + y/x$ now i separate inside the sqrt and => $\displaystyle y' = sqrt(x^2/x^2+y^2/x^2) + y/x$ so now i get => $\displaystyle y' = sqrt(1+(y/x)^2) + y/x$ now i can make sub. $\displaystyle u=y/x => y=ux => y' = u'x + u$ now this becomes => $\displaystyle y' = sqrt(1+u^2) + u$

that's $\displaystyle u'x + u = sqrt(1+u^2) + u$ that's => $\displaystyle u'x = sqrt(1+u^2) + u -u$ => $\displaystyle u'x = sqrt(1+u^2)$ or => $\displaystyle (du/dx)x = sqrt(1+u^2)$ swaping becomes => $\displaystyle (du/sqrt(1+u^2)) = dx/x$ now i integrate both sides, the left one is to be solved with trigonometric substitution and that's $\displaystyle a^2+u^2$ from where i sub. $\displaystyle u = tg0$ (0 theta dono how to write it) now $\displaystyle du = (sec0)^2*d0$ now i get this integral => $\displaystyle Integral(((sec0)^2d0)/sqrt((tg0)^2 +1)$ that's => $\displaystyle Integral(((sec0)^2d0)/sqrt((sec0)^2)$ => $\displaystyle Integral(((sec0)^2d0)/sec0$ => $\displaystyle Integral(sec0d0) = ln|sec0+tg0| + c$ i can write it as $\displaystyle Integral(sec0d0) = ln|sec0+tg0| + lnc1$ cause c is constant which is arbitrary also the lnc1 is constant so now i can join those ln's in one and get => $\displaystyle ln((|sec0+tg0|)*c1))$ now i get this overall => $\displaystyle ln((|sec0+tg0|)*c1)) = ln|x|$ to fix it better i ln both sides and get => $\displaystyle (|sec0+tg0|)*c1) = |x|$ now earlyer i had sub. u = tg0 so now i get this back in place and get $\displaystyle (|sec0+u|)*c1) = |x|$ now the $\displaystyle sec0 = 1/cos0$, from the triangle i know that my hypotenuse is sqrt(1+u^2) hence the sides are 1 and u and now cos0=chatet/hypotenuse => cos0 = 1/sqrt(1+u^2) , also i can find cos0 this way: cause of u = tg0 => u = sin0/cos0 => cos0 = sin0/u now sin0 = oposite/hypotenuse => sin0 = u/sqrt(1+u^2) so cos0 = (u/sqrt(1+u^2)/u) => cos0 = 1/sqrt(1+u^2) from where i get sec0 = sqrt(1+u^2)[/tex] , so my final solution is : $\displaystyle (|sqrt(1+u^2)+u|)*c1) = |x|$ and u = y/x from earlier so => $\displaystyle (|sqrt(1+(y/x)^2)+y/x|)*c1) = |x|$, it can be fixed little more but i think it's ok so far..

2. It would be a lot easier to parse out what you've done if you clean up your LaTeX a bit. Here are some suggestions:

$\displaystyle \sqrt{3}$ is \sqrt{3}.

$\displaystyle \displaystyle\frac{y}{x}$ is \displaystyle\frac{y}{x}.

$\displaystyle \sec(\theta)$ is \sec(\theta).

$\displaystyle \displaystyle\int f(x)\,dx$ is \displaystyle\int f(x)\,dx.

$\displaystyle \ln|x|$ is \ln|x|.

That should help out quite a bit. Also note that you can get some equations to look nicer by including a blank line before and after them. That'll put enough space in there to allow room for various symbols.