# end diff equations,check and express y

• March 5th 2009, 07:37 AM
fiksi
end diff equations,check and express y
(x2 + xy) dy/dx = y2 + xy +x2. Soroban helped me with this one, I get, after substitution and all, y2/x2 + 2(y/x)= 2 logx + c. Is this correct? condition for c is y(1)=1. The problem is, how do I express this properly as y?
And is it right?

second dy/dx=(2x-y)/(x+y).
After sub and all, I get... -1/2log((y2/x2)+(2y/x)-2)=logx +c. cond for c is y(1)=1. Again, can i express this as purely y, and is it correct?

third dy/dx=(2x-y+1)/(x+y). I do sub, w and v, and get:
-1/2log(w2+2w-2)=logu + c. Substituting back, I get... smth like:
1/(x+1/2)2 * e t pwr of c = (y-1/2)2/(x+1/2)2 + 2* (y-1/2)/(x+1/2)-2.
y(1)=1, for c is condition. Now, am i doing this right, and how to express as y,if I can?

So I am stuck somewhere in the end... Integral wizzes I hope can help :) I feel I have messed smth up here...
• March 5th 2009, 08:03 AM
galactus
Quote:

Originally Posted by fiksi
$(x^{2} + xy)y' = y^{2} + xy +x^{2}$. Soroban helped me with this one, I get, after substitution and all, $\frac{y^{2}}{x^{2}} + 2(\frac{y}{x})= 2 log(x) + c$. Is this correct? condition for c is y(1)=1. The problem is, how do I express this properly as y?
And is it right?

I get $C=\frac{-2x^{2}ln(x)+y(2x+y)}{2x^{2}}$

Solving for y, you can express it as $y=x\left(\sqrt{2ln(x)+2C+1}-1\right)$ or $y= -x\left(\sqrt{2ln(x)+2C+1}+1\right)$