Integration only giving one answer?

[Saracen Rue]

Homework Statement

If $a>0$ and $b≠0$, solve the following stating the maximal domain for which the solution is valid:
$\sqrt{a^2-x^2}\cdot \frac{dy}{dx}+b=0,\ y\left(0\right)=0$

Homework Equations

[/B]
$\int _{ }^{ }\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+c,\ a>0$

$\int _{ }^{ }\frac{-1}{\sqrt{a^2-x^2}}dx=\arccos \left(\frac{x}{a}\right)+c,\ a>0$

The Attempt at a Solution

I have no problems doing the actual integration and arrive at the correct solution of $y=-b\arcsin \left(\frac{x}{a}\right),\ \left|x\right|<a\$. However, I also get $y=b\arccos \left(\frac{x}{a}\right)-b,\ \left|x\right|<a\$ as being a valid solution, whereas both the answers and my calculator don't include this solution. Is there a reason as to why this second answer would be omitted?

 

[Mark44]

Homework Statement

If $a>0$ and $b≠0$, solve the following stating the maximal domain for which the solution is valid:
$\sqrt{a^2-x^2}\cdot \frac{dy}{dx}+b=0,\ y\left(0\right)=0$

Homework Equations

[/B]
$\int _{ }^{ }\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+c,\ a>0$

$\int _{ }^{ }\frac{-1}{\sqrt{a^2-x^2}}dx=\arccos \left(\frac{x}{a}\right)+c,\ a>0$

The Attempt at a Solution

I have no problems doing the actual integration and arrive at the correct solution of $y=-b\arcsin \left(\frac{x}{a}\right),\ \left|x\right|<a\$. However, I also get $y=b\arccos \left(\frac{x}{a}\right)-b,\ \left|x\right|<a\$ as being a valid solution, whereas both the answers and my calculator don't include this solution. Is there a reason as to why this second answer would be omitted?

$\arcsin(x) + \arccos(x) = \frac \pi 2$, meaning the two functions differ by a constant, $\pi/2$. Your two answers are different only by a constant.

 

[scottdave]

When you integrate, you must find the constant (c) using the initial conditions. With the correct constants, either arcos or arcsin should work. It looks like you need a constant added

 

[Saracen Rue]

$\arcsin(x) + \arccos(x) = \frac \pi 2$, meaning the two functions differ by a constant, $\pi/2$. Your two answers are different only by a constant.

Ah thank you, I wasn't aware of that relationship. So basically it's similar to saying that $\frac{d}{dx}\left(\tan \left(x\right)\right)=\sec ^2\left(x\right)$ or $\frac{d}{dx}\left(\tan \left(x\right)\right)=\tan ^2\left(x\right)\ +1$.

 

[Ray Vickson]

Homework Statement

If $a>0$ and $b≠0$, solve the following stating the maximal domain for which the solution is valid:
$\sqrt{a^2-x^2}\cdot \frac{dy}{dx}+b=0,\ y\left(0\right)=0$

Homework Equations

[/B]
$\int _{ }^{ }\frac{1}{\sqrt{a^2-x^2}}dx=\arcsin \left(\frac{x}{a}\right)+c,\ a>0$

$\int _{ }^{ }\frac{-1}{\sqrt{a^2-x^2}}dx=\arccos \left(\frac{x}{a}\right)+c,\ a>0$

The Attempt at a Solution

I have no problems doing the actual integration and arrive at the correct solution of $y=-b\arcsin \left(\frac{x}{a}\right),\ \left|x\right|<a\$. However, I also get $y=b\arccos \left(\frac{x}{a}\right)-b,\ \left|x\right|<a\$ as being a valid solution, whereas both the answers and my calculator don't include this solution. Is there a reason as to why this second answer would be omitted?

How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$

 

[scottdave]

How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$

This could work, too. If you think of a right triangle, then arcsin, arccos, and arctan are all returning angles, given a ratio of two sides of the triangle.
So if x represents the adjacent, and a represents the hypotenuse, then arccos(x/a) will return the angle theta, for cos(theta) = x/a
Similarly, x could represent the opposite angle of a triangle, when using arcsin.
So for arctan, you need the opposite/adjacent. Since hypotenuse² = adjacent² + opposite², try working out to see how you could get the ratio for a tangent, using x representing the opposite, and a representing the hypotenuse.

 

[Ray Vickson]

This could work, too. If you think of a right triangle, then arcsin, arccos, and arctan are all returning angles, given a ratio of two sides of the triangle.
So if x represents the adjacent, and a represents the hypotenuse, then arccos(x/a) will return the angle theta, for cos(theta) = x/a
Similarly, x could represent the opposite angle of a triangle, when using arcsin.
So for arctan, you need the opposite/adjacent. Since hypotenuse² = adjacent² + opposite², try working out to see how you could get the ratio for a tangent, using x representing the opposite, and a representing the hypotenuse.

Yes, of course, it is completely elementary. However, I hoped the OP could puzzle it out.

 

[scottdave]

How about a third solution? Maple gets
$$y = -b \arctan \left(\frac{x}{\sqrt{a^2-x^2}} \right) $$

You may find this video about trig integration, interesting. It may shed some light for you.

 

[scottdave]

Yes, of course, it is completely elementary. However, I hoped the OP could puzzle it out.

Whoops, I didn't look closely. I thought the original person had posted that.