Solve this differential equation for the curve & tangent diagram

[so_gr_lo]

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)

y_p - y_m = dy/dx(x_p-x_m)

y_m = 0

y_p = dy/dx(x_p-x_m)

x_m=y_pdy/dx + x_m

x_m is midpoint of OT

x_m = (y_pdy/dx + x_m) /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?

 

[SammyS]

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)

y_p - y_m = dy/dx(x_p-x_m)

y_m = 0

y_p = dy/dx(x_p-x_m)

x_m=y_pdy/dx + x_m

x_m is midpoint of OT

x_m = (y_pdy/dx + x_m) /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 .



Notice that x_p = 2 x_m .

Also, x_p and y_p 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:

 

[STEMucator]

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##?

 

[so_gr_lo]

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.

 

[Mark44]

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)

y_p - y_m = dy/dx(x_p-x_m)

y_m = 0

y_p = dy/dx(x_p-x_m)

x_m=y_pdy/dx + x_m

x_m is midpoint of OT

x_m = (y_pdy/dx + x_m) /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.

 

[STEMucator]

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?

 

[Mark44]

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.

 

[so_gr_lo]

So I think this is the way to do it:

y - y_M = dy/dx(x - x_M)

y_M = 0

y = dy/dx(x - x_M)

x_M = 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.

 

[etotheipi]

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}##.

 

[STEMucator]

I agree with the solution provided by Mark, and eto.

I was providing a hint that related with the link the OP provided.

 

[so_gr_lo]

Ok, I will use the solution given by Mark and eto. Thanks again for the help.

 

[Mark44]

So I think this is the way to do it:

y - y_M = dy/dx(x - x_M)
y_M = 0
y = dy/dx(x - x_M)
x_M = 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:
x_M = 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.