# Math Help - solve the differential...

1. ## solve the differential...

ok so i did this problem and checked in the back of my book and got it wrong..im hoping someone can show me the error of my work.

(y^2+yx)dx - x^2dy = 0

ok so there are a couple techniques that i can use to solve the problem...

1. almost exact
2. homogeneous

im going with homogeneous, because i think its a little easier to use and would require less steps.

so i make my substitution y= ux
therefore dy = udx +xdu

then after i make my substitution..

(u^2x^2 + ux^2) dx - x^2(udx + xdu)

after doing some algebra i got a final answer of

y = xC1 - x/ln|x|

C1 = 1/C

originally i got x/y = C - ln|x| i just flipped it and redefined my C, not sure if i can do this but it felt good? i dunno..lol thanks in advance.

2. after doing some algebra i got a final answer of

y = xC1 - x/ln|x|

C1 = 1/C

originally i got x/y = C - ln|x| i just flipped it and redefined my C, not sure if i can do this but it felt good? i dunno..lol thanks in advance.
Yeah, I think your problem is in here somewhere. x/y = C - ln|x| is correct. Be more careful with your algebra: subtraction and division do not commute!

3. Originally Posted by slapmaxwell1
ok so i did this problem and checked in the back of my book and got it wrong..im hoping someone can show me the error of my work.

(y^2+yx)dx - x^2dy = 0

ok so there are a couple techniques that i can use to solve the problem...

1. almost exact
2. homogeneous

im going with homogeneous, because i think its a little easier to use and would require less steps.

so i make my substitution y= ux
therefore dy = udx +xdu

then after i make my substitution..

(u^2x^2 + ux^2) dx - x^2(udx + xdu)

after doing some algebra i got a final answer of

y = xC1 - x/ln|x|

C1 = 1/C

originally i got x/y = C - ln|x| i just flipped it and redefined my C, not sure if i can do this but it felt good? i dunno..lol thanks in advance.
OK, from here $(u^2x^2+ux^2)dx-x^2(udx+xdu)=0$ we see that

$x^2u^2dx-x^3du=0$ and after dividing by $u^2x^3$ we integrate: $\int\frac{dx}{x}-\int\frac{du}{u^2}=0$

which means that

$\ln|x|+\frac{1}{u}=c$. Now back sub.

4. ok im gonna rework the problem..i was hoping the book might have been wrong..one more quick question..i know i did the problem as homogeneous, but could i have done it as an almost exact differential? find a substitution that would make it exact and get my answer that way? thanks in advance.

5. Originally Posted by slapmaxwell1
ok im gonna rework the problem..i was hoping the book might have been wrong..one more quick question..i know i did the problem as homogeneous, but could i have done it as an almost exact differential? find a substitution that would make it exact and get my answer that way? thanks in advance.
Possibly. I'm not very familiar with that technique, although I do know some DE's succumb to it.