Question

Solve

Answer 1

(1/y) dy=(1/x) dx
what to do after this step

Answer 2

integrate both sides. youll have the natural ln(y)=ln(x)+c then do the exponential of both sides. \[e ^{\ln y}= e^{\ln x +c}\] it'll turn into y=cx

Answer 3

If you want the complete solution, you will have to worry about what values c can take.
Also: is there a constant solution?
What about negative values for x and y?

Answer 4

k,,no i don't need to calcualte C 's value.
But can we solve it like this
ln(y) -ln (x) =C
ln (y/x) =C
e^ln(y/x)=?

Answer 5

I meant: the c in ln(y)=ln(x)+c can be any real number.
If you go on solving for y, you get: \(e^{\ln x+c}=e^c \cdot e^{\ln x}=c_1 \cdot e^{\ln x}=c_1 \cdot x\), where \(c_1\) now is a positive number.

Answer 6

Also, I would do it this way:

\(\ln|y|=\ln|x|+c \Leftrightarrow |y|=c_1|x|, ~~(c_1>0)\).

\(y= \pm c_1|x|=c_2|x|, ~~(c_2 \neq 0)\).

Answer 7

kk