Solve this differential equation for the curve & tangent diagram

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Homework Statement:

Hi,
I have included an attachment of the question about differential equations. I found a similar question and solution on this website:

https://doubtnut.com/question-answer/a-tangent-at-a-point-p-on-the-curve-cuts-the-x-axis-at-a-and-b-is-the-foot-of-perpendicular-from-p-o-262330

Relevant Equations:

Given differential equation:
xdy/dx = 2y
Here is my attempt at a solution:

y = f(x)

yp - ym = dy/dx(xp-xm)

ym = 0

yp = dy/dx(xp-xm)

xm=ypdy/dx + xm

xm is midpoint of OT

xm = (ypdy/dx + xm) /2

Not sure where to go from there because the solution from the link uses with the midpoint of the points A and B intersecting the x-axis, whereas the assignment question concerns the points M and T on the x-axis, where M is the midpoint between the origin and T. Can anybody give any clues?
259E39C1-8D4D-4744-A50A-3E405B0EC5DE.png
 
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Answers and Replies

  • #2
SammyS
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Homework Statement:: Hi,
I have included an attachment of the question about differential equations. I found a similar question and solution on this website:

https://doubtnut.com/question-answe...-is-the-foot-of-perpendicular-from-p-o-262330
Relevant Equations:: Given differential equation:
xdy/dx = 2y

Here is my attempt at a solution:

y = f(x)

yp - ym = dy/dx(xp-xm)

ym = 0

yp = dy/dx(xp-xm)

xm=ypdy/dx + xm

xm is midpoint of OT

xm = (ypdy/dx + xm) /2

Not sure where to go from there because the solution from the link uses with the midpoint of the points A and B intersecting the x-axis, whereas the assignment question concerns the points M and T on the x-axis, where M is the midpoint between the origin and T. Can anybody give any clues?
Hello, @so_gr_lo .

:welcome:

Notice that xp = 2 xm .

Also, xp and yp are the same as x and y, the coordinates of the general point, P .

I have no idea of what the statement regarding points A and B could possibly mean.

Full size image:
259e39c1-8d4d-4744-a50a-3e405b0ec5de-png.png
 
  • #3
STEMucator
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The equation for the tangent line is:

$$y - y_n = m(x - x_n)$$

##m = \frac{dy}{dx}## is the slope, and ##P_n = (x_n, y_n)## is the point. Also note the midpoint formula:

$$(x_n, y_n) = (\frac{x_L + x_H}{2}, \frac{y_L + y_H}{2})$$
$$(x_n, 0) = (\frac{x_L + x_H}{2}, 0)$$

You know ##y_n = 0 \Rightarrow y = \frac{dy}{dx} (x - x_n)## because the midpoint is always on the x-axis for any point ##P_n##. Now ##x_n = \frac{x_L + x_H}{2} \Rightarrow##

$$y + \frac{dy}{dx} (\frac{x_L + x_H}{2}) = \frac{dy}{dx} x$$

What do you know about ##x_L##, and ##x_H##?
 
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  • #4
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x1 = 1/2x2

or

x2 = 1/2x1
I have tried substituting these into y + dy/dx(x1 + x2) = dy/dxx

but can’t see how that helps, I also don’t know how to get rid of on of the dy/dx to turn it into the required differential equation.
 
  • #5
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Homework Statement:: Hi,
I have included an attachment of the question about differential equations. I found a similar question and solution on this website:

https://doubtnut.com/question-answe...-is-the-foot-of-perpendicular-from-p-o-262330
Relevant Equations:: Given differential equation:
xdy/dx = 2y

Here is my attempt at a solution:

y = f(x)

yp - ym = dy/dx(xp-xm)

ym = 0

yp = dy/dx(xp-xm)

xm=ypdy/dx + xm

xm is midpoint of OT

xm = (ypdy/dx + xm) /2

Not sure where to go from there because the solution from the link uses with the midpoint of the points A and B intersecting the x-axis, whereas the assignment question concerns the points M and T on the x-axis, where M is the midpoint between the origin and T. Can anybody give any clues?View attachment 265378
As @SammyS pointed out, you don't need subscripted variables, and the hint given by @STEMucator is much more complicated than is needed. You don't really need either the point-slope form of a line or the midpoint formula.
The proof can be done in two lines, if you assign coordinates to points T and M, and recognize that the derivative dydx gives the slope of the tangent line.
 
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  • #6
STEMucator
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x1 = 1/2x2

or

x2 = 1/2x1
I have tried substituting these into y + dy/dx(x1 + x2) = dy/dxx

but can’t see how that helps, I also don’t know how to get rid of on of the dy/dx to turn it into the required differential equation.
I would double check your math, but please review my prior post. I feel my notation was somewhat unclear before, and maybe the information will help.

Given ##x_L = 0##, and ##x_H = T##, can you create a standard differential equation that you can compute using this hint?
 
  • #7
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I would double check your math, but please review my prior post. I feel my notation was somewhat unclear before, and maybe the information will help.
Your notation was clear enough, but extraneous information such as the point-slope form of the equation of a line and the midpoint formula are overcomplications that are not helpful.

As I said before, all that is needed is to label points P, T, and M in the simplest way possible, and recognize that the slope of the tangent line is ##\frac{dy}{dx}##. When this is done, finding the differential equation requires only two lines.
 
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  • #8
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So I think this is the way to do it:

y - yM = dy/dx(x - xM)

yM = 0

y = dy/dx(x - xM)

xM = 1/2x

y = dy/dx(1/2x)

xdy/dx = 2y

Thanks for the help, since the question had 10 marks I assumed the answer was complicated and got confused by the solution in the link.
 
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  • #9
etotheipi
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Or more simply, the gradient of the tangent line is ##\frac{dy}{dx}##, and that equals the rise over run of that triangle, ##\frac{y}{\frac{x}{2}}##. Namely, ##\frac{dy}{dx} = \frac{2y}{x}##.
 
  • #10
STEMucator
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I agree with the solution provided by Mark, and eto.

I was providing a hint that related with the link the OP provided.
 
  • #11
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Ok, I will use the solution given by Mark and eto. Thanks again for the help.
 
  • #12
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So I think this is the way to do it:

y - yM = dy/dx(x - xM)
yM = 0
y = dy/dx(x - xM)
xM = 1/2x
y = dy/dx(1/2x)
I know what you mean, but that's not what you wrote.
1/2x means ##\frac 1 2 x##, because multiplication and division have the same precedence, and are grouped from left to right. That means that 1/2x is considered to mean (1/2)x, not 1/(2x) as you intended.
To indicate that 2x is in the denominator in both equations above, use parentheses around 2x, like this:
xM = 1/(2x)
y = dy/dx(1/(2x))
so_gr_lo said:
xdy/dx = 2y

Thanks for the help, since the question had 10 marks I assumed the answer was complicated and got confused by the solution in the link.
 
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