Solutions for x<0: Konig's Method

[Konig]

Hi,

I have solved a differential equation (1) for x, where i think is a real variable.

(1) \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1+x)^2}

And get a solution

y(x) = \frac{2}{\sqrt{x}} - \frac{\sqrt{x}}{(1+x)} - 3 \arctan{\sqrt{x}}

Now, i want to find solutions for x<0, and i have said that the above is a solution for x>0.

So far i have simply done a coord transformation x→-x and then solved this equation

\frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2}

and then substituted in "-x" under my square roots.

Is this a correct way of finding x<0?

Konig.

 

[SammyS]

Hi,

I have solved a differential equation (1) for x, where i think is a real variable.

(1) \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1+x)^2}

And get a solution

y(x) = \frac{1}{\sqrt{x}} - \frac{\sqrt{x}}{(1+x)} - 3 \arctan{\sqrt{x}}

Now, i want to find solutions for x<0, and i have said that the above is a solution for x>0.

So far i have simply done a coord transformation x→-x and then solved this equation

\frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2}

and then substituted in "-x" under my square roots.

Is this a correct way of finding x<0?

Konig.

What do you plan to do about the square root of negative numbers?

 

[Konig]

What do you plan to do about the square root of negative numbers?

and then substituted in "-x" under my square roots.

Konig.

So \sqrt{-x}, x&lt;0 will give a positive answer.

Although I am not sure what to do about the x^{\frac{3}{2}} part of the equation.

The answer should have natural logs and some x terms.

the solution of \frac{dy}{dx} = \frac{1}{x^{\frac{3}{2}} (1-x)^2}

gives a similar answer to what i need which gives me hope. But I am not sure what the do with the first terms, eg there signs etc.

thanks.

 

[SammyS]

You can't arbitrarily change the sign of x under the radical.

This will show the problem:

\displaystyle\sqrt{x}=\sqrt{(-1)(-1)x}

\displaystyle =\left(\sqrt{-1}\,\right)\sqrt{-x}​

What will you do with \sqrt{-1}\,? -- even if you call it i .