First order differential equations and constants

[Lord Anoobis]

Homework Statement

$y' = \frac{cos x}{sin y}$

$y' = \frac{6x^2}{y(1 + x^3)}$

Homework Equations

The Attempt at a Solution

So I was working through some textbook problems and there's something about the solutions of the above equations I don't quite understand.
The first one:

$\int sin y\, dy = \int cos x\, dx$
$-cos y = sin x + C$
$y = cos^{-1}(-sin x - C)$
Here the book shows the answer as $cos y = -sin x + C$

The second equation:

With $u = 1 + x^3$ as a substitution we get
$\int y\, dy = 2\int \frac{du}{u}$
$\frac{1}{2}y^2 = 2(\ln |u| + C)$
$y = \pm\sqrt{4(\ln |1 + x^3| + C)}$
Book answer: $y = \pm \sqrt{4\ln|1 + x^3| + C}$

So in the first one the sign of the constant remains positive, in the second it remains as $C$ instead of being multiplied. My question then is: does the $C$ in the book answer simply represent a constant inclusive of its sign and multiple(s) or have I overlooked something?

 

[Ray Vickson]

Homework Statement

$y' = \frac{cos x}{sin y}$

$y' = \frac{6x^2}{y(1 + x^3)}$

Homework Equations

The Attempt at a Solution

So I was working through some textbook problems and there's something about the solutions of the above equations I don't quite understand.
The first one:

$\int sin y\, dy = \int cos x\, dx$
$-cos y = sin x + C$
$y = cos^{-1}(-sin x - C)$
Here the book shows the answer as $cos y = -sin x + C$

The second equation:

With $u = 1 + x^3$ as a substitution we get
$\int y\, dy = 2\int \frac{du}{u}$
$\frac{1}{2}y^2 = 2(\ln |u| + C)$
$y = \pm\sqrt{4(\ln |1 + x^3| + C)}$
Book answer: $y = \pm \sqrt{4\ln|1 + x^3| + C}$

So in the first one the sign of the constant remains positive, in the second it remains as $C$ instead of being multiplied. My question then is: does the $C$ in the book answer simply represent a constant inclusive of its sign and multiple(s) or have I overlooked something?

Different C's, but just denoted by the same letter. Would it help if we give answers to (1) as $\cos y = - \sin x - C_1$ (you) and $\cos y = - \sin x + C_2$ (book)?

BTW: to get properly-typset trig functions in LaTeX, use '\sin x' and '\cos y' instead of 'sin x' and 'cos y'. The same applies to tan, cot, sec, csc and cot plus all the inverse functions arcsin, etc., as well as to exp, log, ln, max, min, lim, etc.

 

[H Smith 94]

For the first one you've got the right answer!

So I was working through some textbook problems and there's something about the solutions of the above equations I don't quite understand.
The first one:

$\int sin y\, dy = \int cos x\, dx$
$-cos y = sin x + C$
$y = cos^{-1}(-sin x - C)$
Here the book shows the answer as $cos y = -sin x + C$

If you apply $\cos$ to both sides, you'll find that you get

\cos{y} = -\sin{x} - c

and since $c$ is an arbitrary constant, we can re-define $C \equiv -c$. Thus:

\begin{equation} \cos{y} = -\sin{x} + C. \end{equation}

The second equation:

With $u = 1 + x^3$ as a substitution we get
$\int y\, dy = 2\int \frac{du}{u}$
$\frac{1}{2}y^2 = 2(\ln |u| + C)$
$y = \pm\sqrt{4(\ln |1 + x^3| + C)}$
Book answer: $y = \pm \sqrt{4\ln|1 + x^3| + C}$

Again, the second question is another matter of re-defining the integration constant. Simply allow $4c \equiv C$ and you'll find that

\begin{equation} y = \pm\sqrt{4\ln\left|1+x^3\right| + C}. \end{equation}

So in the first one the sign of the constant remains positive, in the second it remains as $C$ instead of being multiplied. My question then is: does the $C$ in the book answer simply represent a constant inclusive of its sign and multiple(s) or have I overlooked something?

You seem to be onto the right idea -- the constant is simply a consequence of the integration. Since it is arbitrary, you can define it however you like! It's a handy little tool for cleaning up solutions.

 

[Lord Anoobis]

For the first one you've got the right answer!
If you apply $\cos$ to both sides, you'll find that you get

\cos{y} = -\sin{x} - c

and since $c$ is an arbitrary constant, we can re-define $C \equiv -c$. Thus:

\begin{equation} \cos{y} = -\sin{x} + C. \end{equation}
Again, the second question is another matter of re-defining the integration constant. Simply allow $4c \equiv C$ and you'll find that

\begin{equation} y = \pm\sqrt{4\ln\left|1+x^3\right| + C}. \end{equation}
You seem to be onto the right idea -- the constant is simply a consequence of the integration. Since it is arbitrary, you can define it however you like! It's a handy little tool for cleaning up solutions.

Thanks for clearing that up.

 

[Lord Anoobis]

Different C's, but just denoted by the same letter. Would it help if we give answers to (1) as $\cos y = - \sin x - C_1$ (you) and $\cos y - - \sin x + C_2$ (book)?

BTW: to get properly-typset trig functions in LaTeX, use '\sin x' and '\cos y' instead of 'sin x' and 'cos y'. The same applies to tan, cot, sec, csc ant cot plus all the inverse functions arcsin, etc., as well as to exp, log, ln, max, min, lim, etc.

Thanks for the input, especially that LaTeX tip.