The problem is that find the curve, its slope of tangent lines are twice as steep as the ray from the origin. The point is (x,y)
I solve it like this:(dy/dx)=(2y/x) ∫dy/y=∫dx/x
then ln|y|=2lnx+c (c is constant) then y=Ax^2(A=±e^c)
I think that A is ambiguous up to the constant c, and the graph of the curve I'm aiming for, and the graph of the curves is a bunch of parabolas which is undefined at point (0,0). Did I make any mistakes? If I have any one of the curves to the left of y-axis and to the right connected together and take the derivative of it, can I always get dy/dx=2y/x?
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T^T anyone here?
it sounds right. though the parabola is defined at (0,0)
however, we may want to exclude the origin?
because the "ray from the origin" is undefined at (0,0)
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It's the graph. I can not draw very well.
Oh.. Maybe I can't call it parabolas, it's only similar to parabolas. Because when I solve the differential equations I divided it by x on the right-hand side.
The domain must be restricted to either x<0 or x>0 because the slope is undefined at x=0. From the slope field you can see that as you travel along a particular curve toward 0, the moment you reach 0, you wont know which direction to go because there are too many directions :
are you suggesting the solution is
where A and B are non-zero constants.
Nope, domain depends on the initial value
If the initial value starts negative, we must restrict the domain to x<0 because we wont know which direction that particular solution goes after that singularity at x=0
solution curves are half parabolas
that sounds like philosophy. you could glue two different parabolas together, make sure one of them goes through the initial value, exclude (0,0) and you have a "solution" that meets the conditions.
Do you meas after I take the indefinite integral of the differential equation dy/y=2dx/x I can get a family of functions?
Yes, the solutions are family of half parabolas, not full parabolas with a hole at (0, 0)
If I connect the two curves(any curves, one from the left and one from the right) to gether and get a function then take the derivative of it, can I always get dy/dx=2y/x?
Another question, I think that A is ambiguous up to the constant c, is it right?
Yes for both, \(A=e^C\) is as much an arbitrary constant as \(C\) is
but wait, do we need to worry that A is now positive always ?
NO, because after taking the integral of dy/y we get ln|y|
Ahh right, that does the trick :
\[\ln |y| = \ln|x^2| + C \implies |y| = e^Cx^2 \implies y = \pm e^Cx^2 = Ax^2 \]
\(A\) can be negative too !
My problem is solved now.Thank you! ^ ^