Tag Archives: integrating factors

Study Guide to Bernoulli Equations

Posted on April 30, 2015 by | 1 comment

Overview

What are Bernoulli Equations?

Well to put it simply Bernoulli Equations are first order differential equations. What sets Bernoulli Equations apart from other first order differential equations is that they are nonlinear first order differential equation. When we was first introduced to first order differential equations we learned that the standard form  was :

y’ +p(t)y = g(t) ,  y(to) = yo

What separates Bernoulli Equations from other first order equations is that in standard form, it is not equal to some function that is linear but one that has an exact solution. What this means is that their is some power that is raised to the right side of the equation which we shall call n and n cannot be equal to 0 or 1. this is what makes Bernoulli Equations nonlinear. Now we can change the form of a standard first order linear equation into a nonlinear first order equation:

y’ +p(t)y = q(t)y^(n)  y(to)=yo

The above equation is now the standard form for a Bernoulli equation. What we can now understand from this equation is that  p(t) and q(t) are functions that are continuous. Since n is a real number that has to be n> 0 and n>1 for this  to be non linear.

Now the question is how do we solve Bernoulli Equations?

Sample Problem:

take an equation such as t^(2)y’ + 2ty – y^(3) = 0

and change it by adding y^(3) on both sides and we get:

t^(2)y’ + 2ty = y^(3)

Next we divide by t^(2) to on both sides of the equation so we can get it into standard form.

y’ + 2/yt = y^(3)/t^(2)       <————-STANDARD FORM

Now we can go into the steps to solve this equation;

1. Divide by y^(n)

y’/y^(3) + 2/yt/y^(3) =  (y^(3)/t^(2))/y^(3)

we get: (y’ + 2/yt)/y^(3) = 1/t^(2)

2. We substitute v = y^(1-n)

Since y^(3)

we get:v = y^(1-3) = y^(-2)

now we can rearrange the formula to so we can substitute v into it by moving the y^(2) to the numerator and we get:

y’/y^(3) + 2y^(-2)/t = 1/t^(2)

y’/y^(3) + 2v/t = 1/t^(2) <——————-Substitute v for y^(2)

Before we move to the solve Step we need to take the Derivative of the substitution v = y^(-2)

d/dy y^(-2)

dv/dt = -2y^(-3) dy/dt

(-1/2)v’ = -2y’/y^(3)(-1/2)

Now we can substitute for y’/y^(3) with the above equations and we get:

(-1/2)v’ + (2/t)v = 1/t^(2) <—————– First order linear equation

next we use the method of intergrating factor which states

mu = e^ intergral of (-4/t) which is equal to 1/t^(4)

we then multiply both sides b\by Mu and get the equation:

(dv(t)/dt)/t^(4) + d/dt(1/t^(4))v = (-2/t^(6))

Next we intergrate both sides and we come to the final answer of

y(t) = +- Sqrt(5t)/sqrt(ct^(5) + 2)

Videos:

 

I \\

 

 

 

 

 

1 Comment

Posted in Resources, Study Guide

Tagged , , , , , , ,

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.

 

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

 

 

 

2 Comments

Posted in Resources

Tagged , , , ,