At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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)
|dw:1438353168811:dw| 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 : |dw:1438353913296:dw|
ganeshie8 are you suggesting the solution is Ax^2 x<0 Bx^2 x>0 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.
ganeshie8 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)
ganeshie8 |dw:1438354495749:dw| 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 ?
ganeshie8 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 !
ganeshie8 phi My problem is solved now.Thank you! ^ ^