Can I express the maximal solution interval of an ODE as

[Susanne217]

Homework Statement

I'm working on trying to first to solve an ODE on the form

$$\sqrt{x(t)}\frac{dx}{dt} = 2\cdot t^2$$

My task is to find a solution on the form $$x(t) =nt^s$$

The Attempt at a Solution

I do some separation and I get that the above is equal to

$$\sqrt{x(t)} dx = 2 \cdot t^2 dt$$

Which by integrating on both sides of the equality yields

$$\frac{2}{3} x(t)^{3/2} = \frac{2}{3}t^3$$

I mulitiply by a factor of 3 on both sides and divide by a factor of 2 and arrive at

and get $$x(t)^{3/2} = t^3$$

which gives the solution

$$ln(x(t)) = 2 \cdot ln(t) \Rightarrow x(t) = t^2$$

Then the bad not so easy second question. What is the maximum interval on which x(t) is defined?

According to my textbook I have to equal the two antiderivatives of the original equation to obtain the first endpoint.

and then take the composite of the invers of the anti-derivative of lhs of the original ODE and the antiderivative of the rhs to find the second endpoint.

I then obtain an interval $$I = [t^2, sign(t)]$$

Have I understood this correctly?

 

[HallsofIvy]

I have no idea what you mean by

According to my textbook I have to equal the two antiderivatives of the original equation to obtain the first endpoint.

and then take the composite of the invers of the anti-derivative of lhs of the original ODE and the antiderivative of the rhs to find the second endpoint.

I then obtain an interval
$$I= [t^2, sgn(t)|$$

Equate what two anti-derivatives? The anti-derivatives of each side of your equation $\sqrt{x}dx= 2t^2 dt$? You did that to solve the equation.

Why in the world would you take the composite of the inverses of the two sides? If you have f(y)= g(x), then taking the inverse of f, of both sides, you get $y= f^{-1}(g(x))$ to solve for y but that is the composite of the inverse of the left side with the function on the right side.

And finally, $I= [t^2, sgn(t)]$ means nothing to me. The interval asked for here is the interval between two numbers. Writing two functions together like that is NOT an interval. I would say that because, in the original equation,
$$\sqrt{x}\frac{dx}{dt}= 2t^2$$
the coefficient of the derivative is only defined and differentiable for x> 0, and that is not true if t= 0, the interval of definition is either $(0, \infty)$ or $(-\infty, 0)$, depending on whether an "initial condition" is given for t> 0 or t< 0.

 

[Susanne217]

I have no idea what you mean by

Equate what two anti-derivatives? The anti-derivatives of each side of your equation $\sqrt{x}dx= 2t^2 dt$? You did that to solve the equation.

Why in the world would you take the composite of the inverses of the two sides? If you have f(y)= g(x), then taking the inverse of f, of both sides, you get $y= f^{-1}(g(x))$ to solve for y but that is the composite of the inverse of the left side with the function on the right side.

And finally, $I= [t^2, sgn(t)]$ means nothing to me. The interval asked for here is the interval between two numbers. Writing two functions together like that is NOT an interval. I would say that because, in the original equation,
$$\sqrt{x}\frac{dx}{dt}= 2t^2$$
the coefficient of the derivative is only defined and differentiable for x> 0, and that is not true if t= 0, the interval of definition is either $(0, \infty)$ or $(-\infty, 0)$, depending on whether an "initial condition" is given for t> 0 or t< 0.

Hi Mr Hall,

Thank you for your reply and again I made mistake like a usually do.

I get that the solution to the above ODE is

$$x(t) = (t^3+\frac{3}{2}c)^{\frac{2}{3}}$$

If I draw the solution curve on my graphical Calculator I can see this looks to defined on the interval [-3,3].

Do I then conclude from this Hallsofty that this is the maximum interval for the solution for $$k \geq 1$$??