Question

Find the domain for the particular solution to the differential equation dy/dx = 3y/3x, with initial condition y(1) = 1.

Answer 1

I'm confused about the domain, but I found the separated the variables, took the antiderivatives, and found the particular solution

Answer 2

1/3y dy = 1/x dx
(1/3)ln|x| = ln|x| +C

Answer 3

(1/3)ln|1| = ln|1|+C
0=0+c
0=C

Answer 4

dy/dx = 3y/3x

dy/3y = dx/3x

taking the integral of both sides gives

(1/3)*ln(|y|) = (1/3)*ln(|x|) + C

Multiply everything by 3 to get

(1/3)*ln(|y|) = (1/3)*ln(|x|) + C
3*(1/3)*ln(|y|) = 3*(1/3)*ln(|x|) + 3*C
ln(|y|) = ln(|x|) + C

then solve for y

ln(|y|) = ln(|x|) + C
|y| = e^(ln(|x|) + C)
|y| = e^(ln(|x|)) * e^C
|y| = e^C*e^(ln(|x|))
|y| = e^C*|x|

Now we're given y(1) = 1 which means x = 1 and y = 1

|y| = e^C*|x|
|1| = e^C*|1|
1 = e^C*1
e^C = 1

So the particular solution should be

y = 1*|x|

which simplifies to

y = |x|

For some reason, geogebra is saying the answer is y = x which is a bit odd. I'm not sure why it's not using absolute values

Answer 5

That means the domain is x>0?

Answer 6

That could be a possibility

Answer 7

The answers are
x > 0
x < 0
x ≠ 0
All real numbers

Answer 8

I'm thinking all real numbers because either y = |x| like I showed above or y = x which is what geogebra got. Either way, the domain of each function is the set of all reals

Answer 9

Ok that makes sense, I was thinking more range

Answer 10

Thanks!

Answer 11

glad to be of help