Tag Archives: linear

Homework Hints Week 2: Integrating Factors – A Shortcut

Posted on February 7, 2015 by | 2 comments

Shortcuts are dangerous things – they may save you time, but they usually don’t help you understand the problem.  Because of this, it’s usually important to have a thorough grasp of the basic idea of how to solve a problem before learning the shortcut.  Since you’ve had a few days to wrestle with the “Integrating Factors” problem, I wanted to share a standard shortcut (covered in the text, but not yet discussed in class) for solving these problems, which condenses much of the algebra into two formulas.  You are welcome to use it, or not, as you prefer.

Shortcut for solving Integrating Factors problems:

Step 1:  Rewrite the differential equation in the standard form:  

\frac{dy}{dt} + p(t)y = g(t)

In practice, this usually just means getting the y and \frac{dy}{dt} on the same side, and dividing to get rid of anything in front of the \frac{dy}{dt}.

Step 2:  Find \mu, by plugging in:

\mu = e^{\int p(t) dt}

That is, integrate the function in front of y, and then raise e to the power of the result.  This gives \mu.

Step 3:  Multiply both sides of the equation by \mu, and then integrate both sides.  Notice that the integral of the left side will always equal \mu\cdot y.

Step 4: Finally, solve the resulting equation for y.  You’re done!

Special Bonus Shortcut II:  The work of Steps 3 and 4 can be condensed into the following formula, which can be used to find y directly after completing Step 2:

y = \frac{1}{\mu(t)} \int \mu(t) g(t) dt + C

That is, multiply \mu by the function g(t) from the right hand side of the differential equation, integrate, and multiply the result by $\frac{1}{\mu}$.

NOTE: The standard form mentioned in Step 1 shows up a lot – in fact, even if you are not using the shortcut formulas above, it is considered “pretty standard” to rewrite your equation in standard form before solving the problem.

An example using the Shortcut: NOTE: In the video, he uses x as the independent variable, instead of t.

https://www.youtube.com/watch?v=Et4Y41ZNyao

 

Another example: NOTE: Towards the end of this example, when integrating the right-hand-side, he uses integral of \frac{1}{1+x^2}, which is \arctan x  — if this looks unfamiliar, you should review the derivatives of the inverse trig functions 

 

 

Happy shortcutting,
-Prof Reitz

 

 

 

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