Question

Solve y'=(x+2y)/(2x+y) implicitly.

Answer 1

y=ux
u=y/x
y'=u+u'x
u'x+u=(x+2ux)/(2x+ux)=(x(1+2u))/(x(2+u))=(2u+1)/(u+2)
u'x=(2u+1)/(u+2)-u=(2u+1)/(u+2)-(u(u+2))/(u+2)
=(2u+1-u^2-2u)/(u+2)=(1-u^2)/(u+2)
(u+2)/(1-u^2) du=dx/x

Answer 2

@ganeshie8 @dumbcow

Answer 3

Can you tell me how to integrate from there?

Answer 4

sure, looks good so far

i would use partial fractions to integrate left side
\[\frac{u+2}{(1-u)(1+u)} = \frac{A}{1-u} + \frac{B}{1+u}\]
A - B = 1
A+B = 2

--> A = 3/2, B = 1/2

then after integrating:
\[\rightarrow -\frac{3}{2} \ln (1-u) + \frac{1}{2} \ln (1+u) = \ln x + C\]

Answer 5

And what's next?

Answer 6

you would simply left side, get rid of the logs

then you have to sub "y/x" in for u to get final answer

if possible you may try to solve for "u" or "y" explicitly but it may not be possible and the question says implicitly so its not possible for this problem

Answer 7

How do I simplify that? I know that I need to multiply (1-u)(1+u) but what about the -3/2 and 1/2?

Answer 8

remember your log properties

n*log(x) = log(x^n)

log(a) - log(b) = log(a/b)

to make it easier i would first multiply everything by 2 to get rid of fractions