How do I know it's the positive branch

[thomas49th]

dy/dx = (4x - y)/(x + 2y)

for y(0) = 1

As not seperable us y = vx where v is a funciton of x method.

If you follow it through you'll get y as a quadratic. According to the answer scheme, I should take the +ve square root because of y(0) = 1... why? Why can I discard the -ve square root.

Thanks
Thomas

 

[Mark44]

dy/dx = (4x - y)/(x + 2y)

for y(0) = 1

As not seperable us y = vx where v is a funciton of x method.

If you follow it through you'll get y as a quadratic. According to the answer scheme, I should take the +ve square root because of y(0) = 1... why? Why can I discard the -ve square root.

Thanks
Thomas

Your solution is defined in some interval around x = 0. Since y(0) = 1, y(x) for x near 0 should be close to 1, hence positive.

 

[thomas49th]

Sorry I don't follow. The only link I see if 1 is +ve... If y(0) = -1 would I choose -ve branch? Simply because -1 is negative?

 

[vela]

Yeah, you can't satisfy the initial condition (in the original problem) if you choose the negative solution.

 

[thomas49th]

Yeah, you can't satisfy the initial condition (in the original problem) if you choose the negative solution.

Hmmm. I set up an equation where 1 equalled the quadratic formulae with the negative branch. If you work through it, you find the discrement is < 0, therefore meaning no real solutions. Is what you mean by no being able to satisfy the initial condition... no real solutions?

However, from the answer sheet, it seems they did it intuitively... how?

Thanks
Thomas

 

[Mark44]

Hmmm. I set up an equation where 1 equalled the quadratic formulae with the negative branch. If you work through it, you find the discrement is < 0, therefore meaning no real solutions. Is what you mean by no being able to satisfy the initial condition... no real solutions?

This doesn't have anything to do with the quadratic formula or the discriminant. When you were finding a solution, you had two choices for y - a positive square root or a negative square root. The one to choose is determined by the initial condition. Since y(0) = 1 is a positive number, choose the positive root.

If the initial condition had been y(0) = -2, you should have chosen the negative root.

That's all there is to it.

However, from the answer sheet, it seems they did it intuitively... how?

Thanks
Thomas