# Thread: Differential equation for straight lines through origin

1. ## Differential equation for straight lines through origin

Sketch a few members of the straight lines through the origin. By eliminating parameters, derive a differential equation which describes this family.

Attempt

$\displaystyle y = mx$ (since there are infinite gradients)

$\displaystyle \frac{dy}{dx} = m$

but the correct answer was: $\displaystyle \frac{dy}{dx} = \frac{y}{x}$. I am not sure why.

2. Well, your answer "dy/dx= m" isn't correct because you have not eliminated the parameter m!

From the original equation, y= mx, m= y/x which gives dy/dx= y/x.

3. Thanks. If instead the question was: find the differential equation for the straight lines through the point (2,3), what would the procedure be?

I tried using:

y = mx + c
3 = 2m + c

but differentiating it obviously wouldnt work, since there are only constants

4. Originally Posted by SyNtHeSiS Thanks. If instead the question was: find the differential equation for the straight lines through the point (2,3), what would the procedure be?

I tried using:

y = mx + c .... (1)
3 = 2m + c .... (2)

but differentiating it obviously wouldnt work, since there are only constants
I have numbered the equations. Solve for c in terms of m from (2). Substitute this expression for c into (1). Now answer the question.

5. ## Re: Differential equation for straight lines through origin

y=mx+c
Since, c=0
m=y/x
Differentiating on both sides....
0=y dx - x dy / y²

1 dx/y - x dy/y² =0

1 dx/y = x dy/ y²

1 dx / x = 1 dy / y

y/x = dy/dx

And that's your required answer ��

,

,

,

,

,

,

,

,

,

,

,

,

,

,

# find the differential equation of all straight lines at a unit distance from the origin

Click on a term to search for related topics.

#### Search Tags

differential, equation, lines, origin, straight 