# Differential Equation

1. Sep 18, 2009

### KillerZ

1. The problem statement, all variables and given/known data

Verify that the indicated function $$y = \Phi(x)$$ is an explicit solution of the given first-order differential equation. Give at least one interval I of definition.

2. Relevant equations

$$2y^{'} = y^{3}cos x$$
$$y = (1 - sin x)^{-1/2}$$

3. The attempt at a solution

I think I did the first part right but I am not sure about the interval I of definition.

$$y^{'} = -\frac{1}{2}(1 - sin x)^{-3/2}(- cos x)$$

Left hand side:
$$2y^{'} = 2(-\frac{1}{2}(1 - sin x)^{-3/2}(- cos x))$$
$$= -(1 - sin x)^{-3/2}(- cos x)$$
$$= (1 - sin x)^{-3/2}(cos x)$$

Right hand side:
$$y^{3}cos x = ((1 - sin x)^{-1/2})^{3}(cos x)$$
$$= (1 - sin x)^{-3/2}(cos x)$$

Therefore $$y = (1 - sin x)^{-1/2}$$ is a solution.

$$I = (-\infty, \pi/2)$$ or $$(\pi/2, \infty)$$ is the interval I of definition.

2. Sep 18, 2009

### LCKurtz

Apparently you have noted that sin(x) can't be 1. But x = pi/2 isn't the only place that happens. You only have to give an interval and it will have to be shorter.

3. Sep 18, 2009

### KillerZ

Something like this?

$$I = (-3\pi/2, \pi/2)$$ or $$(\pi/2, 2\pi/4)$$

4. Sep 18, 2009

### LCKurtz

Yes and no. You must have a typo in the second one.

5. Sep 18, 2009

### KillerZ

Yes its:
$$I = (-3\pi/2, \pi/2)$$ or $$(\pi/2, 5\pi/2)$$

Thank You