Intro. to Differential Equations

hawaiidude

hey thanks...nice examples,,,clear and easy to understand...but heres 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

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 differnt 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 differnt 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

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 lets 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

Awsome, I feel special now that this was made a sticky [6)]

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

My understanding is that PF postings will support latex formatting. I haven't used it yet but go to: http://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

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