huclogin one year ago 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?

T^T anyone here?

2. phi

it sounds right. though the parabola is defined at (0,0) however, we may want to exclude the origin?

3. phi

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.

6. ganeshie8

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|

7. phi

ganeshie8 are you suggesting the solution is Ax^2 x<0 Bx^2 x>0 where A and B are non-zero constants.

8. ganeshie8

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

9. ganeshie8

solution curves are half parabolas

10. phi

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?

12. ganeshie8

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?

14. ganeshie8

Yes for both, $$A=e^C$$ is as much an arbitrary constant as $$C$$ is

15. ganeshie8

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|

17. ganeshie8

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! ^ ^

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