Question

Find the condition that the constants a and b must satisfy in order that y = ax + b satisfies the equation:

x(dy/dx)^2 - y(dy/dx) = 2

Find the two solutions of this equation satisfying y = 3 when x = 2.

Would appreciate a walk through as I'm working through this for revision, esp. at the initial re-arranging stage.

Answer 1

a y'^2 - y y' = 2 is the question? this is differential equation stuff then right?

Answer 2

Yes, x y'^2 - y y' = 2

Answer 3

cause it could also be quadratic stuff

Answer 4

differential stuff

Answer 5

x(y'^2) -y(y') -2 = 0

Answer 6

if y=3 and x=2 we get what:

2.-2 = -4

4 - 1 = 3

(y' +4)(y'-1) = 0 is what i come up with

Answer 7

opps forgot to divide out the 2 :)

Answer 8

(y' +2) (2y'-1) = 0

Answer 9

... but i got know idea if this is helpful

Answer 10

Those are the solutions, but how can I find the constraints?

Answer 11

\[\frac{dy}{dx} = \frac{d}{dx}(ax + b) = \frac{d}{dx}ax+\frac{d}{dx}b=a+0=a.\]
\begin{eqnarray*}x \left(\frac{dy}{dx}\right)^2 - y\left(\frac{dy}{dx}\right) &=&xa^2 - (ax + b)a \\&=&xa^2 - a^2x-ab\\&=&-ab=2.\end{eqnarray*}
If \[y(3) = 2,\]\[y(3) = 3a + b = 2 \Rightarrow b = 2-3a \Rightarrow a(2-3a) = -2\]
\[\Rightarrow 3a^2 -2a - 2 = 0 \Rightarrow a = \frac{1}{3}\left(1 \pm \sqrt{7}\right) \land b = 1 \mp \sqrt{7}.\]

Answer 12

gonna have to rely on one of them there smarter than mes for that one ;)

Answer 13

so in general should I sub in a for y'

Answer 14

No, that's only because the problem said that\[y = ax+b.\]If we didn't have that information, the problem would be much more difficult.

Answer 15

Thanks for the help, is there any way of saving threads? Just found this openstudy thing this evening, really going to help with my revision.

Answer 16

your questions are saved on your profile; just click on your screenname

Answer 17

Sweet, thanks