Intro. to Differential Equations

[hawaiidude]

lol no it is correct

 

[Integral]

Originally posted by hawaiidude
ok here's another one solve ...
:: .
x -6x+9x=y

how would u solve this

Where are the differentials? You need to make this a differential equation. As it stands I get

y=4x

Not real exicting.

 

[ExtravagantDreams]

Originally posted by Integral
Where are the differentials? You need to make this a differential equation. As it stands I get

y=4x

Not real exicting.

Yeah, that's what I was thinking.

 

[ExtravagantDreams]

Repeated Roots; Reduction of Order

This is a method for finding a second solution to a 2nd order linear homogeneous differential equations with constant coefficients assuming you already have the first solution. It can often occur when the roots of the equation are the same (when b² - 4ac = 0).

Lets remember that given the equation;

ay'' + by' + cy = 0,

where a, b, and c are constant coefficients,

There can be a solution found by first finding the roots of;

ar² + br + c = 0

Then it is a simple matter of remember the equations;

C₁e^r₁t + C₂e^r₂t

for ordinary roots, and;

C₁e^λtcos(μt) + C₂e^λtsin(μt)

for complex roots in the form;

r = λ ± iμ

The general idea is to find a non-linear multiple of the first solution. Given the first solution, y₁ = e^rt, the general solution is C₁e^rt and another solution y₂ = v(t)y₁, where v(t) is some funtion of t, can be found.

Using;
y₂ = v(t)y₁

Find;
y'₂ = v'(t)y₁ + v(t)y'₁
y''₂ = v''(t)y₁ + 2v'(t)y'₁ + v(t)y''₁

And plug these into the original equation;

ay'' + by' + cy = 0

a[v''(t)y₁ + 2v'(t)y'₁ + v(t)y''₁] + b[v'(t)y₁ + v(t)y'₁] + c[v(t)y₁]

Now group the equation such that all v(t), v'(t), and v''(t) terms are together and if it was done correctly, all v(t) terms will cancel.

v''(t)[ay₁] + v'(t)[2ay'₁ + by₁] + v(t)[ay''₁ + by'₁ + cv(t)y₁]

Since,
b² - 4ac = 0

ay''₁ + by'₁ + cv(t)y₁ = 0

So you are left with;

v''(t)[ay₁] + v'(t)[2ay'₁ + by₁]

Solve for v(t) by integrating and plug back into;
y₂ = v(t)y₁

Let's try an example:

y'' + 4y' + 4y = 0

Find the Roots;
r₁₂ = [-4 ± √(4² - 4(1)(4))]/2(1)
r₁₂ = -2

y₁ = e^-2t

y₂ = v(t)e^-2t
y'₂ = v'(t)e^-2t - 2v(t)e^-2t
y''₂ = v''(t)e^-2t - 4v'(t)e^-2t + 4v(t)e^-2t

Plug into original equation;
v''(t)e^-2t - 4v'(t)e^-2t + 4v(t)e^-2t + 4[v'(t)e^-2t - 2v(t)e^-2t] + 4[v(t)e^-2t]

Combine v(t) terms;
v''(t)[e^-2t] + v'(t)[-4e^-2t + 4e^-2t] + v'(t)[4e^-2t - 8e^-2t + 4e^-2t] = 0

v''(t)[e^-2t] + v'(t)[ 0] + v'(t)[ 0] = 0

In this case the v'(t) terms also cancelled. So,

v''(t)[e^-2t] = 0

Since,
e^-2t [x=] 0,
v''(t) = 0

Integrating;
v'(t) = C₁,
v(t) = C₁t + C₂,

Remember the original solution;
y₁ = e^-2t

and;
y₂ = v(t)e^-2t

y₂ = [C₁t + C₂]e^-2t
y₂ = C₁te^-2t + C₂e^-2t

Where the second term is just the the first, so the general solution is;
y = C₁te^-2t + C₂e^-2t


It actually turns out that y₂ = ty₁, always.

Give these problems a try:

1. y'' - 2y' + y = 0
2. 4y'' + 12y' + 9y = 0

 

[hawaiidude]

hey thanks...nice examples,,,clear and easy to understand...but here's a problem...when x=0 3x^2y''-xy'+y=0

 

[ExtravagantDreams]

I'm not sure what you are asking;

3x²y''- xy'+ y = 0,

When x = 0

3(0)²y''- (0)y'+ y = 0

y = 0 ?

 

[hawaiidude]

not really but how do you find recurrence formulas? they're very complicated and i can't understand it...like the recurrence for (x^2+4)y''+xy=x+2

i thought you find the seond derivative and and first and the original

a0 )a1x+a2x^2+a3x^3+a4x^4...+anx^n+an+1+an+2x^n+2+...

y'=a1+2azx+3a3x^2...and so on..

how would you compute this? iam very confused ...all i got is you get the combining terms in this case8a2=2, 24a3+a0=1 2a2+48a4+a1=0 ...

then it's like n(n-1)an+4(n+2)(n+1)an+2+an-1=0 (n=0, 1 ,2 ,3 ,4 ...x

i know how they got the combined terms but how did they get the n's?

 

[ExtravagantDreams]

Are you talking about the sums in series solutsions of 2nd Order Linear Equations? If it is beyond that, sorry I can't help ya. I'm only taking this course just now.

 

[ExtravagantDreams]

Nonhomogeneous Equations; Method of Undetermined Coefficients

Here we will look at Second Order, nonhomogeneous Linear Equations of the form;

L[y] = y'' + p(t)y' + q(t)y = g(t)

where p(t), q(t), and g(t) are continuous functions on an open interval, I. We can use the homogeneous equations, where g(t) = 0 to solve the nonhomogeneous.

If Y₁ and Y₂ are two solutions of the nonhomogeneous equation, then their difference (Y₁ - Y₂) is a solution of the corresponding homogeneous equation. If in addition, y₁ and y₂ are a fundamental set of solutions of the homogeneous equation, then;

Y₁ - Y₂ = c₁y₁(t) + c₂y₂(t)

where c₁and c₂ are certain constants.


The general solution of the nonhomogeneous equation can then be written as;

y = φ(t) = c₁y₁(t) + c₂y₂(t) + Y(t)

where y₁ and y₂ are a fundamental set of solutions of the corresponding homogeneous equation, c₁ and c₂ are arbitrary constants, and Y(t) is some specific solutions of the nonhomogeneous equation.

We will attempt here to find the function Y(t) of the nonhomogeneous equation. There are two methods of doing this, namely the method of undetermined coefficients, which is discussed in this section, and the method of variation of parameters, which will be discussed next time.

The idea is to assume a solution for Y(t) with an undetermined coefficient, then use this answer and plug back into the original equation and try to find the coefficient. If it is successfuly, a solution of Y(t) has been found, if not there is so solution in the form that was assumed and a different assumption should be made.

Clearly, this has draw back, such that an assumption must be fairly easy to do. Yet, once such an assumption has been made, the solution is not difficult to optain.

Let's look at some examples:

y'' - 3y' - 4y = 3e^2t

Here we seek a function such that the combination of Y''(t) - 3Y'(t) - 4Y(t) = 3e^2t

Since, the exponential function reproduces itself through differentiation it is the most plausible answer. Let's assume Y(t) = Ae^2t, where A is the undetermined coefficient.

Y(t) = Ae^2t
Y'(t) = 2Ae^2t
Y''(t) = 4Ae^2t

Plug these values into the combination equation;

1[4Ae^2t] - 3[2Ae^2t] - 4[Ae^2t] = 3e^2t

Attempt to solve for A;

[4 - 6 - 4]A = 3
A = -1/2

So;

Y(t) = -1/2e^2t

Let us try another one, where we first assume the incorrect solution;

y'' - 3y' - 4y = 2sin(t)

Let us assume;

Y(t) = Asint(t)
Y'(t) = Acos(t)
Y''(t) = -Asint(t)

1[-Asint(t)] - 3[Acos(t)] - 4[Asint(t)] = 2sin(t)

A[-5 - 3cot(t)] = 2

Clearly, this can not be solved. Let's assume a different solution, namely Y(t) = Asint(t) + Bcos(t), where B is just another undetermined coefficient.

Y(t) = Asint(t) + Bcos(t)
Y'(t) = Acost(t) - Bsin(t)
Y(''t) = -Asint(t) - Bcos(t)

1[-Asint(t) - Bcos(t)] - 3[Acost(t) - Bsin(t)] - 4[Asint(t) + Bcos(t)] = 2sin(t)

[-A + 3B -4A]sin(t) + [-B - 3A - 4B]cos(t) = 2sin(t)

Since there there are two sin(t) on the right there must be two on the left, zero cos(t) on the right, there must be zero on the left.

Hence the coefficients of sin(t) and cos(t) must be;

-5A + 3B = 2,
-3A - 5B = 0

1 -3/5 -2/5
0 -34/5 -6/5

1 0 -5/17
0 1 3/17

A = -5/17
B = 3/17

So, Y(t) = (-5/17)sint(t) + (3/17)cos(t)

Lets try one more,

y'' - 3y' - 4y = 4t² - 1

Since g(t) is a polynomial with terms t², t, 1, with coefficients 4, 0, -1 respectively, we can assume the solution;

Y(t) = At² + Bt + C
Y'(t) = 2At + B
Y''(t) = 2A

[2A] - 3[2At + B] - 4[At² + Bt + C] = 4t² - 1

[-4A]t² + [-6A - 4B]t + [2A - 3B - 4C] = 4t² - 1

-4At² = 4t²
A = -1

-6At - 4Bt = 0t
-6(-1) = 4B
B = 3/2

2A - 3B - 4C = -1
2(-1) - 3(3/2) - 4C = -1
-4C = 11/2
C = -11/8

So, Y(t) = (-1)t² + (3/2)t + (-11/8)

Problems:

Find the general solution of;
1. y'' - 2y' - 3y = 3e^2t
2. 2y'' + 3y' + y = t² + 3sin(t)

3. y'' + 2y' + 5y = 4e^-tcos(2t), y(0) = 1, y'(0) = 0

 

[hawaiidude]

yeah thanks...by the way, how do you compute pde's? the advanced types?

 

[ExtravagantDreams]

I believe that is Fourier series, something I will not learn until Monday.

 

[hawaiidude]

i thought pde's were partil differential equations>?

 

[selfAdjoint]

He meant that's one way to solve them, I think. Yes PDE means partial differential equation. Some of them can be reduced to ordinary differentials and there are just tons of methods for particular cases. This has been one of the most active branches of math resarch for hundreds of years, and no end in sight.

 

[hawaiidude]

o..by the way..here are some things that i wish to know
wha are the recurrence formulass for the folowwing

1) (x^2 +4) y''+xy=x+2
2) y''+y=0
8x^2 y''+10xy'+(X-1)y=0

 

[ExtravagantDreams]

Nonhomogeneous Equations; Variation of Parameters

We have already seen how to find a particular solution for nonhomogeneous equation using the Method of Undetermined Coefficients, now we will try to use variation of parameters to accomplish the same thing.

Let's jump straight to an example;

y'' + 4y = 3csc(t)

Noting that the corresponding homogeneous equation is;

y'' + 4y = 0

We first solve this equation.

r₁₂ = √(-4(1)(4))/2(1) = 2i

remember that solution will be in form;

e^λt[cos(μt) + sin(μt)]

where in this case λ = 0 and μ = 2, so

y_c(t) = c₁cos(2t) + c₂sin(2t)

The basic idea is to replace the constance c₁ and c₂ with functions u₁(t) and u₂(t) and solve for these functions.

Starting with the equation;

y = u₁(t)cos(2t) + u₂(t)sin(2t)

we can differentiate to optain;

y' = u'₁(t)cos(2t) + u'₂(t)sin(2t) - 2u₁(t)sin(2t) + 2u₂(t)cos(2t)

Since we only have one initial condition so far, yet two unknown variables, this would give us infinite many solutions. Let us impose a second condition so that we have one final solution. Here it is not important why we can do this;

We require that;

u'₁(t)cos(2t) + u'₂(t)sin(2t) = 0, so;

y' = 2u₂(t)cos(2t) - 2u₁(t)sin(2t)

y'' = 2u'₂(t)cos(2t) - 2u'₁(t)sin(2t) - 4u₂(t)sin(2t) - 4u₁(t)cos(2t)

Substitude these equations back into the original equation;

[2u'₂(t)cos(2t) - 2u'₁(t)sin(2t) - 4u₂(t)sin(2t) - 4u₁(t)cos(2t)] + 4[u₁(t)cos(2t) + u₂(t)sin(2t)] = 3csc(t)

2u'₂(t)cos(2t) - 2u'₁(t)sin(2t) = 3csc(t)


From our second set condition;
u'₁(t)cos(2t) + u'₂(t)sin(2t) = 0

u'₂(t) = -u'₁(t)cos(2t)/sin(2t)


Substitude;
2[-u'₁(t)cos(2t)/sin(2t)]cos(2t) - 2u'₁(t)sin(2t) = 3csc(t)

Simplify;

u'₁(t) = -(3csc(t)sin(2t))/2 = -3cos(t)

Substituging once more;

u'₂(t) = -u'₁(t)cos(2t)/sin(2t)
u'₂(t) = [-3cos(t)]cos(2t)/sin(2t)
u'₂(t) = (3/2)csc(t) -3sin(t)

Now that we have optained u'₁(t) and u'₂(t), Integrate;

u₁(t) = -3sin(T) + c₁
u₂(t) = (3/2)ln|csc(t) - cot(t)| + 3cot(t) + c₂(t)

Finally, substitude u₁(t) and u₂(t) into the y expression;

y = [-3sin(T) + c₁]cos(2t) + [(3/2)ln|csc(t) - cot(t)| + 3cot(t) + c₂(t)]sin(2t)


That probably looked more confusing than need be, so let's look at an arbitrary function to see a set by set method and prove this can be used for any Second Order Linear Nonhomogeneous Equation.

Let us start with the general equation;
y'' + p(t)y' + q(t)y = g(t)

The general solution to the corresponding homogeneous equation will be;
y_c(t) = c₁y₁(t) + c₂y₂(t)

This is from the assumption that the equation has constant coefficients. Now in the general solution, replace constants with functions u.

y = u₁(t)y₁(t) + u₂(t)y₂(t)

Take the derivative;
y' = u'₁(t)y₁(t) + u'₂(t)y₂(t) + u₁(t)y'₁(t) + u₂(t)y'₂(t)

For a second condition set terms with u' equal to zero;
u'₁(t)y₁(t) + u'₂(t)y₂(t) = 0

This gives;
y' = u₁(t)y'₁(t) + u₂(t)y'₂(t)

Differentiate again and plug y, y', and y'' into the original equation;

y'' = u₁(t)y''₁(t) + u₂(t)y''₂(t) + u'₁(t)y'₁(t) + u'₂(t)y'₂(t)

[u₁(t)y''₁(t) + u₂(t)y''₂(t) + u'₁(t)y'₁(t) + u'₂(t)y'₂(t)] + p(t)[u₁(t)y'₁(t) + u₂(t)y'₂(t)] + q(t)[u₁(t)y₁(t) + u₂(t)y₂(t)] = g(t)

Rearranging;
u₁(t)[y''₁(t) + p(t)y'₁(t) + q(t)y₁(t)] + u₂(t)[y''₂(t) + p(t)y'₂(t) + q(t)y₂(t)] + u'₁(t)y'₁(t) + u'₂(t)y'₂(t)] = g(t)

Since both y₁ and y₂ are solutions to the corresponding homogeneous equation, the expressions in brackets equal zero, leaving;

u'₁(t)y'₁(t) + u'₂(t)y'₂(t) = g(t)

Using this equation and the previous equation;
u'₁(t)y'₁(t) + u'₂(t)y'₂(t) = 0

substituation can be used and integration can be done to find u₁and u₂.

u'₁(t) = -y₂(t)g(t)/W(y₁,y₂)(t)

u'₂(t) = y₁(t)g(t)/W(y₁,y₂)(t)

u₁(t) = -∫(y₂(t)g(t)/W(y₁,y₂)(t))dt + c₁

u₂(t) = ∫(y₁(t)g(t)/W(y₁,y₂)(t))dt + c₂

Where then;
Y(t) =
-y₁(t)∫(y₂(t)g(t)/W(y₁,y₂)(t))dt + y₂∫(y₁(t)g(t)/W(y₁,y₂)(t))dt

and the general solution is;
y = c₁y₁(t) + c₂y₂(t) + Y(t)

I realize this is a little confusing to follow, so let me sum up what you really need to know without deriving everything everytime.

First;
You must find the solutions y₁ and y₂ of the homogeneous equation.

Then use these two formulas:

u'₁y₁ + u'₂y₂ = 0
u'₁y'₁ + u'₂y'₂ = g(t)

If there is a term infront of the y'' term, it must be divided out to give the correct g(t).

From this system of equations, where y and y' are known, u'₁ and u'₂ can be found, then integrated.

 

[QuantumNet]

DDy + Dy + y = 0 for all values of y

x² + px + q cannot be zero for all values.

Although it's possible if y = ae^kx.

y = ae^kx
Dy = ake^kx.
DDy = ak²e^kx.

DDy + Dy + y = ak²e^kx + ake^kx + ae^kx = a(k² + k + 1)e^kx

If (k² + k + 1) = 0 then DDy + Dy + y = 0

 

[ExtravagantDreams]

Series solution of 2nd Order Linear Equations: Ordinary Point

Awsome, I feel special now that this was made a sticky

I haven't been around in a while, I was really busy during finals and then kinda crawled in a hole for a month during break. but for now I am back, we will see how long it lasts this time.

I changed my format of doing this, instead of writing everything out on here, I have decided to use Word Processor and Equation Editor to create a document that you can then download. I hope no one will have any problems this way.

If you dare...

 

[nickdanger]

However, a question, does anyone know an easier way for writing math on the computer and one that looks less confusing. I know I will have difficulty finding some things, especially subscripts and superscripts. Anyone know a better way to denote these?

My understanding is that PF postings will support latex formatting. I haven't used it yet but go to: https://www.physicsforums.com/showthread.php?t=8997
for instructions.

 

[Ebolamonk3y]

Does anyone know of any good Intro to Diffy Q books? Or just Diffy Q books in general? Thanks...

 

[abacus]

Does anyone know of any good Intro to Diffy Q books? Or just Diffy Q books in general? Thanks...

I'm using the book Differential Equations: 2nd Edition by Blanchard, Devaney, Hall.

 

[Dr Transport]

Try Boyce and DiPrima...it hasn't been thru 7 or 8 editions beause it is not a good, readable text

 

[quantumdude]

Try Boyce and DiPrima...it hasn't been thru 7 or 8 editions beause it is not a good, readable text

I second that. I actually took the Diff Eq with Boyce--great teacher.

The Schaum's outline "Modern Introductory Differential Equations" is also very good.

 

[tavi_boada]

Apostol's calculus is great

 

[zeronem]

Im going to begin by implicit differentiation, then I will ask the question I have. The question is on finding the interval where solution is valid.

xy^2 - e^{-y} - 1 = 0

u = x, v = y

(u + du)(v+dv)^2 = (u + du)(v^2 + 2vdv + d^2v)

= uv^2 + 2uvdv + ud^2v + v^2du + 2vdvdu + d^2vdu

we get rid of uv^2 and any part of the equation that contains more then one differential. Let's say w = uv^2, then w + dw = the above equation. dw would equal the above equation without the uv^2 since you subtract the w. You get rid of the parts of the equation that has more then one differential because more then one differential is just simply too small to have any effect on the whole equation. We then get

2uvdv + v^2du

We then plug in for x and y and get

2xyy' + y^2

We then have implicitly differentiated part of the equation. We now continue to implicitly differentiate the rest of the equation.

z = -y

-e^{-y} = -e^z

dz = -dy

\frac {d(-e^z)} {dz} = -e^z = -e^{-y}

we then take dz and multiply it by \frac{d(-e^z)}{dz} ,

(dz)(-e^{-y}) = + e^{-y}y'

we then differentiate -1 which becomes a 0 and we get

2xyy' + e^{-y}y' + y^2 = 0

As implicit differentiation of xy^2 - e^{-y} - 1 = 0

We then test the implicit differentiation of the above equation as the solution to the differential equation

(xy^2 + 2xy - 1)y' + y^2 = 0

We do this by solving for e^{-y} from the equation xy^2 - e^{-y} - 1 = 0

e^{-y} = xy^2 - 1

we plug e^{-y} in the implicit differential equation and get (2xy + xy^2 -1)y' + y^2 = 0

So it is a solution of the differential equation (xy^2 + 2xy - 1)y' +y^2 = 0

Now this is where I need help. Can I say that the original equation xy^2 - e^{-y} - 1 = 0 is an implicit solution to (xy^2 + 2xy -1)y' + y^2, even though it works out?


Exactly how do I find such interval where the y is an implicit function of x on an interval. So I can determine whether the equation xy^2 - e^{-y} - 1 = 0 is an implicit solution to (xy^2 + 2xy -1)y' + y^2

 

[Fritz]

<br /> \frac {dy} {dt} = ay - b<br />

Here is y the independent variable and a and by the dependent variables?

 

[QuantumTheory]

Help !

You get rid of the parts of the equation that has more then one differential because more then one differential is just simply too small to have any effect on the whole equation.1 = 0 [/tex] is an implicit solution to ..

YES, YES, YES!

This is a problem I am having!

I understand that a lot of people here dislike the book Calculus For Dummies, for beginners. a lot of people in the real world cannot understand calculus, it is hard. This book really opened my eyes. I understood that a derivative if nothing but a constant rate. I learned the basics to move on.

However, before I got this book, I picked up my first calculus book in my school library. It was made in the very early 1900's, and was about calculus. Beginner calculus.

In this book, he actually had equations of multi-varibles!

For example, dx^2/dy^2, or anything with an infinitely small piece of a deriviative squared.

Since a deriviative is an infinitely small section, out of a whole of an infinite amount of infinitely small pieces, how could you have basically a negative infinity squared? This just doesn't make sense. You're already at some negative infinity rate of one dx. So if the derivative is the rate of some curve(therefore the height of the curve), let's call it f(t).

Then,

f(t) = f(t)dx


Does this make sense? I made it up myself..I hope its right.

Basically, I don't understand how you could possibly have some derivative, of some function, squared.

Ugh, anyway, I hate infinity, it confuses me sometimes :(

Weird how an infinite amount of infinitely thin pieces can equal a finite amount, eh?

 

[poolwin2001]

The link is not working.

 

[whozum]

I have to say this guide is amazing. I have one question on post 25:

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b ± √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t

How did you get from the above to the last statements?

 

[dextercioby]

Which one,the first statement?

Daniel.

 

[whozum]

How does:

Assuming that these roots are real and different then;
y1(t) = er1t
y2(t) = er2t

follow from

py'' + qy' + ry = 0

Solving the homogeneous equation will later always provide a way to solve the corresponding nonhomogeneous problem.

I'm not going to proove all this but you can take the kernal of this funtion as

ar2 + br + c = 0

and you can, so to speak, find the roots of this funtion.

r1,2 = (-b ± √(b2 -4ac))/2a

r1 = (-b + √(b2 -4ac))/2a
r2 = (-b - √(b2 -4ac))/2a

I think i understand the root analogy and how r1 and r2 come about, but I don't understand why y1 and y2 follow that structure.