Question

find the general solution of the equation y'=(x^2+3y^2)/(2xy)

Answer 1

i get y = e^(3x^2/8)

Answer 2

can anyone verify this?

Answer 3

wolfram disagrees.

Answer 4

It would take me too long to look up how to do it though I'm afraid. I need to review diff-eqs.

Answer 5

\[y'=\frac{x^2+3y^2}{2xy}\]

\[2xy y'=x^2+3y^2\]

\[2xyy'-3y^2=x^2\]

\[x \cdot 2yy'-3y^2=x^2\]

------------------------
Recall (y^2)'=2yy'

\[2yy'-3\frac{1}{x} y^2=x\]
assuming \[x \neq 0\]

now we need to multiply v on both sides such that we can write
\[(vy^2)'\]
on the right hand side
so how do we choose this v
well let's think
how can we get
v 2yy'+v' y^2 on right hand side

we need v'=-3/x

so v is a fuction of x so we can write
\[\frac{dv}{dx}=\frac{-3}{x}\]

we can use separation of variables to find what v needs to be

\[dv= \frac{-3}{x} dx\]

integrate both sides

\[v=-3 \ln|x|+C_1\]
let's just use 0 for C_1

so we have

\[v=-3 \ln|x|\]

so we have

\[-3 \ln|x| 2y y'+ \frac{-3}{x} y^2 =-3x \ln|x|\]
so cleaning this up a bit we have
\[-3 \ln|x| (y^2)'+(-3 \ln|x|)'y^2=-3x \ln|x|\]

so now we can write

\[(-3 \ln|x| y^2)'=-3 x \ln|x|\]

now integrate both sides

\[-3 \ln|x| y^2+K_1=-3 (\frac{x^2}{2} \ln|x|-\int\limits_{}^{} \frac{x^2}{2} \frac{1}{x} dx +K_2)\]
\[-3 \ln|x| y^2+K_1=-3(\frac{x^2}{2} \ln|x|-\frac{1}{2} \cdot \frac{x^2}{2}+K_2)\]
Note: \[-3K_2-K_1=K\]


\[-3\ln|x|y^2=\frac{-3x^2}{x}\ln|x|+\frac{3x^2}{4}+K\]

Answer 6

does wolfram agree with that?

Answer 7

wolfram just said it was x sqrt (x-1) + c

Answer 8

but i have no clue how to get anywhere near that

Answer 9

we can see what happens if we tried so solve for y

Answer 10

oops that was a mistake on the last line that x is suppose to be a 2 on the bottom of that one fraction

Answer 11

\[-3\ln|x|y^2=\frac{-3x^2}{2} \ln|x|+\frac{3x^2}{4}+K\]

Answer 12

\[y^2=\frac{x^2}{2}+\frac{-x^2}{4 \ln|x|}+\frac{K}{-3 \ln|x|}\]


hmmm....
i don't this will end up looking like what they got

Answer 13

:(

Answer 14

i did make a mistake
i will redo ok?

Answer 15

i owe you a huge favor for this one myinin

Answer 16

:)

Answer 17

no favors needed