First order differential equations and constants

In summary: It would definitely help if the solutions were denoted by different constants, as that would make it clearer that they are just arbitrary constants and can be re-defined as needed.
  • #1
Lord Anoobis
131
22

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?
 
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  • #2
Lord Anoobis said:

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.
 
Last edited:
  • #3
For the first one you've got the right answer!

Lord Anoobis said:
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

[tex] \cos{y} = -\sin{x} - c [/tex]

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

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

Lord Anoobis said:
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}

Lord Anoobis said:
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.
 
Last edited:
  • #4
H Smith 94 said:
For the first one you've got the right answer!
If you apply ##\cos## to both sides, you'll find that you get

[tex] \cos{y} = -\sin{x} - c [/tex]

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.
 
  • #5
Ray Vickson said:
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.
 
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1. What is a first order differential equation?

A first order differential equation is an equation that involves the first derivative of a function with respect to one independent variable. It can be written in the form dy/dx = f(x), where y is the dependent variable and x is the independent variable. Solving a first order differential equation involves finding the function y(x) that satisfies the equation.

2. How do constants play a role in first order differential equations?

Constants in first order differential equations are typically used to represent initial conditions or physical constants in the equation. They can also be used as arbitrary values to simplify the equation. The values of the constants can affect the behavior and solutions of the equation.

3. What methods can be used to solve first order differential equations?

There are several methods for solving first order differential equations, including separation of variables, integrating factors, and using the power series method. The choice of method depends on the form of the equation and the initial conditions given.

4. How are first order differential equations used in science and engineering?

First order differential equations are used in various fields of science and engineering to model and analyze systems that involve change over time. They are particularly useful in areas such as physics, chemistry, biology, and economics to describe and predict the behavior of dynamic systems.

5. What are some real-life applications of first order differential equations?

Some common applications of first order differential equations include modeling population growth, radioactive decay, chemical reactions, and electrical circuits. They are also used in fields such as fluid mechanics, heat transfer, and economics to describe the rates of change in physical quantities.

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