Separable Differential Equations

Posted by Jeff Schuler on Jul 27, 2009 in Uncategorized |

Separable Differential Equations

These are 1st order differential equations, they can be written in the form y^I = f(t)g(y)

Why is it called a first order equation?

You can tell the order of any equation by looking at the highest power of the differential (\frac{\text{d}}{\text{d}x}) on either side of the equation,  the below case, we have a FIRST order differential equation because (\frac{\text{d}}{\text{d}x}) is raised to the power of 1.

\frac{\text{d}y}{\text{d}x} - y^2 e^{x} = \cos(x) + e^{x} + \cos(x) y^2

Our goal is to solve for y.  But how can we do that if we have a differential in our equation?  In order to start, we need to isolate the \frac{\text{d}y}{\text{d}x} .

\frac{\text{d}y}{\text{d}x} = \cos(x) + e^{x} + \cos(x) y^2 + y^2 e^{x}

Combine any like-terms that you can, this will make things way easier and you will struggle to finish one of these without doing this.  As you can see below, I have found e^{x} occurring more than once on the right side, so I can pull that out as a common factor.

\frac{\text{d}y}{\text{d}x} = \cos(x) + \cos(x) y^2 + e^{x} (1 + y^2)

We are not done yet, I see one more, \cos(x)

\frac{\text{d}y}{\text{d}x} = \cos(x)(1 + y^2) + e^{x} (1 + y^2)

Things are starting to look a little bit neater, but we can take this 1 step further, (1 + y^2) is a like-term as well, we now have this.

\frac{\text{d}y}{\text{d}x} = (1 + y^2)(\cos(x) + e^{x})

Note: y^{2} + 1 = 0 has no real solutions (square root of a negative number is imaginary)

Alright, what should we do from here?

Grouping Like Terms

We need to get all of the ‘x’ terms together and all of the ‘y’ terms together, this includes the dy and dx on the left side.  As you can see, \frac{\text{d}y}{\text{d}x} is simply a fraction, multiply the dx to the right side of the equation to end up with this.

dy = (1 + y^2)(\cos(x) + e^{x})dx Almost done, now we move all terms with ‘y’ in them to the left side like so,

\frac{\text{d}y}{(1 + y^2)} = (\cos(x) + e^{x})dx

Now we have something we are familiar with!  Using integration tables, TI-89, or just plain skills, you should now integrate each side of the equation so that you can have these derivatives (dy and dx) as simply variables (x & y).  This is simple calculus.

\int_{}^{} \frac{1}{(1 + y^2)}(dy) = \int_{}^{} \cos(x) + e^{x}(dx)

Which equates to

\arctan(y) = \sin(x) + e^{x} + C y = \tan(\sin(x) + e^{x} + C)

What is “C”?!

The Integration Constant

The “C” represents a constant that is generated via the integration process.  When you take a derivative of a constant, the result is Zero, this must be accounted for when working backwards with an integral.  This is calculus as well.  In later lessons, we will be manipulating this constant and solving for it, in order to do this, we need initial conditions given to us (or some professors will try to be tricky by making you solve for those as well).  This constant is often represented with a “K”, as well.

We have succeeded in finding y as a function of x, we started with a very cryptic equation that included differentials, y’s and x’s jumbled up together.  As you can see, through a series of combining like-terms,  variable manipulation and integration, one can solve for an unknown variable.

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