Please someone help me with this "differential equation" problem.

Prove that for any straight line through origin $\frac{y}{x} = \frac{dy}{dx}$. and interpret this.

Question

Please someone help me with this "differential equation" problem. 

Prove that for any straight line through origin $\frac{y}{x} = \frac{dy}{dx}$. and interpret this.

lochana · Answer

should I integrate it?

dls · Answer

Its simply saying that slope of a line = m = y/x = tan theta (if it passes through origin then c=0 in the standard equation of line y=mx+c)

lochana · Answer

I see.. but do you know how to prove it?

dls · Answer

$$\Large \frac{y}{x} = \frac{dy}{dx}$$
$$\large \int\limits_0^y \frac{dy}{y} = \int\limits_0^x \frac{dx}{x}$$

lochana · Answer

yeah. then it becomes$lnx = lny$.. now I can say it is y=x. right?

dls · Answer

You missed a step. Even though its obvious but let me state it.
$$\Large \ln(y) = \ln(x) + c$$
Since it passes through origin then (0,0) satisfies it. Hence c=0
Now you can simply take antilog on both the sides and get y=x :)

lochana · Answer

got it. @DLS Thanks for showing it. it really helped. bye

baru · Answer

Y=cx (where c is the constant of integration)

ganeshie8 · Answer

basically the differential equation $dy/dx = y/x$ represents the family of straight lines passing thru origin

ganeshie8 · Answer

$y = cx\tag{1}$

differentiating both sides with respect to x gives

$dy/dx =c$

from $(1)$ we have $c = y/x$

baru · Answer

or if you want to integrate,
ln y =ln(x) +ln(c)
ln(y)=ln(x*c)
y=cx