I am doing some work on integration and differential equations. I got stumped by the following problems.

Q1: Show that the substitution y = vx, where v is a function of x, transforms the differential equation
xy(dy/dx) = x ² + y²
into the differential equation vx(dv/dx) = 1.

Q2: The normal at every point on a given curve passes through the point (2,3). Prove that the eqn of the curve satisfies the differential eqn:
(x-2) + (dy/dx)(y-3) = 0

Hence find the cartesian eqn of the family of curves which satisfy the condition above. (I think I can do this part)

Find the eqn of the perpendicular curve which passes through the point (4,2).

Hal,

If y=vx, we have dy/dx=v+x(dv/dx). Substituting, we have

x(vx)(v+x(dv/dx))=x² +(vx)²

Now cancel the common factor x² and a little bit more rearranging gives you the required form.

Now, for the second question. Do you know the general formula for the equation of normal? The equation of normal is

(dy/dx)_(x0,y0) (y-y₀ )+(x-x₀ )=0

And for the very last bit, two curves y=y₁ (x) and y=y₂ (x) are perpendicular if and only if

(dy₁ /dx)(dy₂ /dx)=-1

at the point(s) of intersection.

Hope this helps.

Kerwin

Hi Kerwin,

Thanks for the quick response.

The problem for me to solving Q1 was that I was not sure how exactly to tackle y=vx. I see that you have done this : (dy/dx)= v + x(dv/dx). I however don't completely understand how you have implicitly differentiated this. Why does the 'vx' part become 'v + x(dv/dx)'. Is this becaise v is a function of x. I think this part would be better understood if I fully comprehended the meaning of the line 'v is a function of x'. Am I right in thinking that x is a function of y, thats why you get the LHS of y ---> (dy/dx), and v is a function of x, so vx are a product so it becomes x/(dv/dx). I am I on the right line of thinking?

The second part of the Q1 is posing me more of a problem than I initially thought. As the original differential eqn is xy(dy/dx)= x ² + y ² , I am not fully aware of what to do with the with any of the sides to get the x's and y's to either side of the eqn. It got very messy when I tried rearranging it. Here is how far I got:

Originally we have: xy(dy/dx) = x ² + y²
Then lets try to: y(dy/x) = x + (y ² /x)
And divide through by y ² : y ^-1 (dy/dx) = (x/y) + (1/x)
Then (1/y)(1-x)(dy/dx) = (1/x)

Regarding question 2:
The general form that you showed I think I know in the form, (note m=gradient), so (y-y₁ ) = (1/m)(x-x₁ ), then rearrange you get m₁ (y-y₁ ) -(x-x₁ ).

But how does that answer the question in proving it passes the point (2,3). Even though the point initially takes was (2,0). How do you put it into writing as an answer.

I did not get the right answer in the cartesian form either.

Regards
HAL2001

Kerwin

Thanks Kerwin!

Its much clearer now.
The trick was not to assume that v was a constant and diff wrt x. And treat the vx with the product rule.

The second part of the question, the reason I got thrown off the path was that the question said to use the original differential eqn, thus my goofed up attempt. The eqn you used works.

Thanks once again.
HAL2001

Hal,

Now for the second question, the normal to the curve has gradient -(dx/dy) evaluated at x=x₀ and y=y₀ ., where (x₀ ,y₀ ) is the point on the curve.

so we have, assuming y=y₁ (x), gradient of normal is -1/(dy₁ /dx)_(x0,y0) passing through the point (x₀ ,y₀ ) on the curve. So we have

(y-y₀ )=-1/(dy₁ /dx)_(x0,y0) (x-x₀ )

Rearranging gives (dy₁ /dx)_(x0,y0) (y-y₀ )+(x-x₀ )=0.

Since the normal passes through (2,3), we have

(dy₁ /dx)_x0,y0 (3-y₀ )+(2-x₀ )=0

Now, since this properites is true for all point on the curve, we can replace x₀ by x and y₀ by y, giving the required answer as y₁ =y.

Kerwin

Thanks for that further explanation Kerwin.

I don't quite understand why the last line of the post turned out like (3-y ₀ )+(2-x ₀ ), I thought that it would be the other way round like, m ₂ (y-3) + (x-2)=0

I also got confused with the notation in using subscript 1s and 0s generally. What does the line 'assuming y=y₁ (x)' mean?

I think I get the general drift of the solution that you proposed. Though it has not quite completly hit me to be instantly obvious, yet.

Regards
HAL2001

Hal,

Well, the last line should read

(dy₁ /dx)_(x0,y0) (3-y₀ )+(2-x₀ )=0,

which when multiplied both side by -1 and replacing x₀ by x, y₀ and y₁ by y gives the required solution.

The reason I used y₁ (x) is completely arbitrary. You can use simply y(x) if you wanted. Using y₁ (x) has the advantage on the second part, when I can use y₂ (x) for the curve normal to y₁ (x).

Hope this explains it.

Kerwin

Thanks Kerwin,

I see what you mean now.
The m1 x m2=-1.

Once again, thanks for your help.
HAL2001