Intro. to Differential Equations

StephenPrivitera

Here's my solution to the first one:
dy/dx=x²/y
y(dy/dx)=x²
[inte] y(dy/dx)dx = [inte] x²dx
(1/2)y²= (1/3)x³ + C₁
y²= (2/3)x³ + C₂

ExtravagantDreams

"I've often wondered why this is the case. dy and dx are variables right? So why can't they be treated as such?"

x is a variable, y is a funtion of x. By cross multiplying you would be treating y like a variable.

And your answer is correct.

ExtravagantDreams

I will mention right now, I have some difficulty understanding this part so I will have even more difficulty explaining it and will probably need a little help. The good news, this isn't the most important topic in my opinion.

The reason this is applicable is that it is often easier to find the existence and uniquness of a solution without having to work out the actual problem.

Theorm:
Let the function f and &part;f/&part;y be continuous in some rectangle [alpha] < t < [beta], &Gamma; < y < &delta; containing the point (t₀, y₀). Then, in some interval t₀ - h < t < t₀ + h contained in [alpha] < t < [beta], there is a unique solution y = &phi;(t) of the initial value problem
y' = f(t,y), y(t₀) = y₀


Assuming that both the function and it's partial deivative with respect to y are continuous in a rectangle R; |t - t₀| < A, |y - y₀| < B;

Let:
M = max |f(t,y)|, (t,y) is in R
C = min (A,B/M)

Then:
There is one and only one solution y(t) valid for
t₀- C < x < t₀+ C



Ex.
y' = y², y(0) = 1
f(t,y) = y²
&part;f/&part;y = 2y

M = max |f(t,y)|, (t,y) is in R
C = min (A,B/M)

Since A can be made infinately large find the max for B

M = max|(1 + B)²|
C = B/M

C = B/(1 + B)²
C = 1/4

If you have no idea where the answer came from, you are in the same boat as I. If you do know, please share.

Perhaps another example

Ex. 2
y' = (t² +y²)^3/2, y(t₀) = y₀
&part;f/&part;y = 3y(t² + y²)^1/2

M = max|(t² +y²)^3/2|
M = [(t₀ + A)² + (y₀ + B)²]^3/2

C = min(A,B/M)
C = min(A, B/[(t₀ + A)² + (y₀ + B)²]^3/2)


Also, does anyone know how to display determinants or matracies on the computer?

Integral

Keep an eye on what you are doing in the above examples. You are finding a specific region upon which you can guarantee that a unique solution exists. This means you must provide some specific numbers to the bounds of the region.

One thing that might help you a little, redraw your picture with the A and B intervals CENTERED on (t₀,y₀)

In your worked example the interval for y is:
B-y₀<=y <= B+y₀

Since y₀=1 we have:
1-B<=y<=B+1

so if M= Max|f(t,y)| and f(t,y)= y² The maximum value of f(t,y) on our interval is (B+1)²
I believe that this is saying you are free to pick any value on the t axis for t₀, once that value is defined, the value of f(t[₀,y) will be fixed and you can guarantee the existence of a Max value for f(t,y)On an interval surrounding it.This follows from the definitions.
To get this a value of B must be given or defined, since you are free to choose B to be anything, it looks like someone picked 1.

Does this help?

ExtravagantDreams

I guess I am getting confused because the book does not use
M = max |f(t,y)|
C = min (A,B/M)

notation. From what I understand, all this really says is that you are trying to find the maximum interval of f(t,y), where the limiting factor is either A or B/M (which is denoted C). But why B/M, why not just B? This is what gets me. Is C a value in the horizontal direction? Does dividing B by M make it a horizontal value? And how do you choose B?

I think I am making this too difficult. I fear it is one of those things where you just have to look at it.

ExtravagantDreams

Theorem:
Let the functions M, N, M_y, N_x, where subscripts denote partial derivatives, be continuous in the rectangular region R: [alpha] < [beta], &Gamma; < y < &delta;. Then;

M(x,y) + N(x,y)y' = 0

is an exact differential equation in R if and only if

M_y(x,y) = N_x(x,y)

at each point of R. That is, there exists a function satisfying

&part;[psi]/&part;x(x,y) = M(x,y), &part;[psi]/&part;y(x,y) = N(x,y)
or
[psi]_x(x,y) = M(x,y), [psi]_y(x,y) = N(x,y)

if and only if M and N satisfy

M_y(x,y) = N_x(x,y)


This means that there is a solution [psi](x,y) of the general equation M + Ny' = 0

Proof of M_y(x,y) = N_x(x,y)
We already defined

[psi]_x(x,y) = M(x,y), [psi]_y(x,y) = N(x,y)

we can compute the partial derivative of each to be

[psi]_xy(x,y) = M_y(x,y), [psi]_yx(x,y) = N_x(x,y)

Since M_y and N_x are continuous, it follows that [psi]_xy and [psi]_yx are continuous also, which also guarantees their equality.

Finding [psi](x,y)
when M_y(x,y) = N_x(x,y):
Starting with the equations

[psi]_x(x,y) = M(x,y), [psi]_y(x,y) = N(x,y)

start with the first and integrate with respect to x

[psi](x,y) = &int;M(x,y)dx + h(y)

Where h is some function of y playing the role of an arbitrary constant. Now we must proove that h(y) can always be chosed so that [psi]_y = N

[psi]_y(x,y) = &part;/&part;y[&int;M(x,y)dx + h(y)]
[psi]_y(x,y) = &int;M_y(x,y)dx + h'(y)

Setting [psi]_y = N we obtain

N(x,y) = &int;M_y(x,y)dx + h'(y)

Where we can then solve for h'(y)

h'(y) = N(x,y) - &int;M_y(x,y)dx
h(y) = N_x(x,y) - M_y(x,y)

Then the general solution

[psi](x,y) = &int;M(x,y)dx + &int;[N(x,y) - &int;M_y(x,y)dx]dy


Ex. 1

2xy³ + 3x²y²y' = 0

Where M = 2xy³, N = 3x²y²
Then M_y = 6xy², N_x = 6xy²

Since M_y = N_x
This equations is exact and can be solved by the previous method.

Start with

[psi]_x = M = 2xy³
[psi] = &int;Mdx + h(y) = &int;2xy³dx + h(y)
[psi] = x²y³ + h(y)

Find h;
Since we know that N = &part;[psi]/&part;y, differentiate both sides and substitude N

N = &part;/&part;y[&int;Mdx + h(y)
N = 3x²y² + h'(y)

h'(y) = 0, h(y) = C

Plug h(y) back into the original [psi] equation
[psi] = x²y³ + C

[psi] = x²y³ = K

Ex. 2
Find the function [psi](x,y) of
y' = -(ax + by)/(bx - cy)

(bx - cy)y' = -(ax + by)
(ax + by) + (bx - cy)y' = 0

So
M = (ax + by), N = (bx - cy)
M_y = b, N_x = b

Since M_y = N_x, the equation is exact

[psi]_x = M = ax + by
&int;[psi]_xdx = &int;(ax + by)dx
[psi] = (1/2)ax² + byx + h(y)
[psi]_y = &part;/&part;y[(1/2)ax² + byx + h(y)]
[psi]_y = bx + h'(y)

[psi]_y= N

N = bx + h'(Y)

(bx - cy) = bx + h'(y)

h'(y) = -cy
h(y) = -(1/2)cy²

[psi] = (1/2)ax² + byx - (1/2)cy² = K


Ex. 3
(ycosx + 2xe^y) + (sinx + x²e^y - 1)y' = 0

So,
M = ycosx + 2xe^y, N = sinx + x²e^y - 1
M_y = cosx + 2xe^y, N_x = cosx + 2xe^y

M_y = N_x, so can be solved using the exact method

Remember that
M = [psi]_x
N = [psi]_y

&int;[psi]_xdx = &int;(ycosx + 2xe^y)dx
[psi] = ysinx + e^yx² + h(y)

Taking a partial derivative of both sides with respect to y,

[psi]_y = sinx + x²e^y + h'(y)

[psi]_y = N = sinx + x²e^y - 1

sinx + x²e^y - 1 = sinx + x²e^y + h'(y)
h'(y) = -1
&int;h'(y)dy = h(y) = -y

[psi](x,y) = ysinx + e^yx² - y = k


Integrating Factor
If an equation is not exact it can be multyplied by an integrating factor [mu] so that it becomes exact.

Starting with the general form
M(x,y) + N(x,y)y' = 0

and
M_y [x=] N_x

There is an integrating factor [mu] such that
[mu]M(x,y) + [mu]N(x,y)y' = 0

and
([mu]M)_y = ([mu]N)_x

or

[mu]_yM + [mu]M_y = [mu]_xN + [mu]N_x
M[mu]_y - N[mu]_x + (M_y - N_x)[mu] = 0

We will not discuss finding [mu](x,y) since this is entirely too diffictuly for this course, so we shall stick to [mu] as a function of x or y only.

If [mu] is a function of x;
([mu]M)_y = [mu]M_y, ([mu]N)_x = [mu]N_x + N(d[mu]/dx)

Thus, if ([mu]M)_yis to equal [mu]N_x;
d[mu]\dx = (M_y - N_x)[mu]/N

If the function (M_y - N_x)/N depends on x only then [mu] can also be a function of x only and an integrating factor has been found.

The proceedure to finding [mu](y) is similar and the equation
d[mu]\dx = -(M_y = N_x)[mu]/M
is derived.

Ex. 4
(3xy + y²) + (x² + xy)y' = 0
M = 3xy + y², N = x² + xy
M_y = 3x + 2y, N_x = 2x + y

Since,
M_y [x=] N_x

the equation is not seperable and an integrating factor must be found

[mu](x)
(M_y - N_x)/N dx = d[mu](x)/[mu](x)
ln|[mu](x)| = &int;(3x + 2y - 2x - y)/(x² + xy)dx
ln|[mu](x)| = &int;(x + y)/x(x + y)dx
ln|[mu](x)| = &int;(1/x)dx
ln|[mu](x)| + ln|x|
[mu](x) = x

Since it is a function of x only, this can be used as the integrating factor. Now to solve the equations

x(3xy + y²) + x(x² + xy)y' = 0
M = 3x²y + xy², N = x³ + x²y
M_y = 3x² + 2xy, N_x = 3x² + 2yx

M_y = N_x
So it is now exact.

&part;/&part;y[&int;(3x²y + xy²)dx] = x³ + x²y
&part;/&part;y[yx³ + (1/2)y²x² + h(y)] = x³ + x²y
x³ + x²y + h'(y) = x³ + x²y

h'(y) = 0, h(y) = C

yx³ + (1/2)y²x² = K

Problems:

1. (2x + 3) + (2y - 2)y' = 0
2. (9x² + y - 1)dx - (4y - x)dy = 0, y(1) = 0
3. (x²y = 2xy + y³)dx + (x² + y²)dy = 0

Integral

This is covered pretty well in my copy of B&D.

He is using Picard iterations to prove the existance of solutions. I'll cheat for this cause he says it better then I. Here is are links to scans of Boyce & Deprima 2nd Ed.First page second page
Hope this helps, is this not in your book?

ExtravagantDreams

A second order differential equation is linear if it is in the following form;
y'' + p(t)y' + q(t)y = g(t)
or
P(t)y'' + Q(t)y' + R(t)y = g(t)

Where p, q, and g and functions of t

If the problem has initial conditions it will be in the form of
y(t) = y₀, y'(t) = y'₀.

Let's start with the easiest:
Homogeneous with Constant coefficients

Homogeneous in this case means g(t) = 0,
with constant coefficients mean P(t) = p, Q(t) = q, and R(t) = r
Since makes the general equations;

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.

r_1,2 = (-b &plusmn; &radic;(b² -4ac))/2a

r₁ = (-b + &radic;(b² -4ac))/2a
r₂ = (-b - &radic;(b² -4ac))/2a

Assuming that these roots are real and different then;
y₁(t) = e^r₁t
y₂(t) = e^r₂t

and y = C₁y₁(t) + C₂y₂(t)

Which comes from the initial deriviation. If someone really really wants to know, I will show you.

y = C₁e^r₁t + C₂e^r₂t

This is your general solution.
C₁ and C₂ can be solved for if initial conditions y(t) and y'(t) are given in the following manner;

y = C₁e^r₁t + C₂e^r₂t

y(t) = y₀

y₀ = C₁e^r₁t + C₂e^r₂t

and

y' = r₁C₁e^r₁t + r₂C₂e^r₂t


y'(t) = y'₀


It is also possible to varify your solution by using
y = C₁e^r₁t + C₂e^r₂t
y' = r₁C₁e^r₁t + r₂C₂e^r₂t
y'' = r₁²C₁e^r₁t + r₂²C₂e^r₂t

and pluging them back into the equation
ay'' + by' + cy = 0

Ex. 1
y'' + 5y' +6y = 0, y(0) = 2, y'(0) = 3

1r² + 5r + 6 = 0

(r + 3)(r + 2)

r₁ = -3
r₂ = -2

y = C₁e^-3t + C₂e^-2t

2 = C₁e^-3(0) + C₂e^-2(0)
2 = C₁ + C₂


y' = -3C₁e^-3t - 2C₂e^-2t

3 = -3C₁e^-3(0) - 2C₂e^-2(0)
3 = -3C₁ - 2C₂

|+1 +1 +2|
|- 3 - 2 +3| ~

|+1 +1 +2|
|+0 +1 +9| ~

|+1 +0 - 7|
|+0 +1 +9| ~

C₁ = -7
C₂ = 9

y = -7e^-3t + 9e^-2t
y' = 21e^-3t - 18e^-2t
y'' = -63e^-3t + 36e^-2t

y'' + 5y' +6y = 0
(-63e^-3t + 36e^-2t) +5(21e^-3t -18e^-2t) =6(-7e^-3t + 9e^-2t)

e^-3t(105 - 63 - 42) + e^-2t(36 - 18 - 18) = 0

hawaiidude

ok heres question.... solve the differential equation d^3x/dt^3-2d^2d^2x/dt^2 +dx/dt=0



and find the fourier series for f(x)={0,pi -pi<x<0
0<x<pi

ExtravagantDreams

I'm actually not entirely sure how to solve this. This is not something we covered (higher order linear equations)

Skimming through the section, it looks like it is done in the same way as regular constant coefficient homogeneous equations.

3r³ - 2r² + 1r = 0

where r are the kernals

r(3r² - 2r + 1) = 0

r₁ = 0
r₂₃ = (1 &plusmn; i&radic;(2))/3


y = C₁ + e^{1/3 x}(C₂cos((&radic;(2)/3)x) + C₃sin((&radic;(2)/3)x))

and I'm not sure what a fourier series is.

ExtravagantDreams

I have fallen way behind in doing this, about 3 + 3 fairly complicated sections, so I will try to cover as much as I can during my time off tonight and tomorrow morning.

So, we already saw what happens to 2nd order linear homogeneous differential equations with constant coefficient (wow, that is a mouth full) in the form;

ay'' + by' + cy = 0

where one finds the roots by evaluating;

ar² + br + c = 0

and the answer is in the form;

y = C₁e^r₁t + C₂e^r₂t

However, when the expression

b² - 4ac

is negative, the roots can be written in the form;

r₁₂ = &lambda; &plusmn; i&mu;

Where;
&lambda; = -b/2a
i&mu = &radic;(b² - 4ac)
when b² < 4ac

and

y₁(t) = e^{(&lambda; + i&mu;)t}
y₂(t) = e^{(&lambda; - i&mu;)t}

Let us look at some properties first:

Using taylor serious it can be prooven that
e^it = cos(t) + isin(t)
e^-it = cos(t) - isin(t)
e^i&mu;t = cos(&mu;t) + isin(&mu;t)

Where then
e^{(&lambda; + i&mu;)t} = e^&lambda;te^i&mu;t = e^&lambda;t[cos(&mu;t) + isin(&mu;t)]

Looking for real solutions

Addition of function 1 and 2
y₁(t) + y₂(t) =
(e^&lambda;t[cos(&mu;t) + isin(&mu;t)]) + (e^&lambda;t[cos(&mu;t) - isin(&mu;t)]) = 2e^&lambda;t[cos(&mu;t)]

Subtraction of function 1 and 2
y₁(t) - y₂(t) =
(e^&lambda;t[cos(&mu;t) + isin(&mu;t)]) - (e^&lambda;t[cos(&mu;t) - isin(&mu;t)]) = 2e^&lambda;t[isin(&mu;t)]

Since the differential equation is made up of real coefficients, it can be said that it's derivative is also. Therefore by simply neglecting the constant multipliers we obtain a pair of real-valued solution;


u(t) = e^&lambda;t[cos(&mu;t)]
v(t) = e^&lambda;t[sin(&mu;t)]

where u and v are just the real and imaginary parts of the solution respectively; meaning these parts are linearly independent, and a combination of these parts is a then also a solution.

W(u,v)(t) [x=] 0

Since the Wronskian of these two functions is equal to &mu;e^2&lambda;t, it is always nonzero as long as &mu; [x=] 0. Since &mu; is always greater then zero, when b² < 4ac, the general solution is;

y = C₁e^&lambda;tcos(&mu;t) + C₂e^&lambda;tsin(&mu;t)

Ex.
y'' + y' + y = 0

r² + r + 1

r₁₂ = (-1 &plusmn; &radic;[(1)² - 4(1)(1)])/2(1) =
-1/2 &plusmn; i&radic;(3)/2

Then;
&mu; = -1/2
&lambda; = &radic;(3)/2

y = C₁e^-t/2cos(&radic;(3)t/2) + C₂e^-t/2sin(&radic;(3)t/2)

hawaiidude

ok heres another one solve ......
:: .
x -6x+9x=y

how would u solve this

ExtravagantDreams

LOL, are you trying to make fun of me?

What is the solution to the previous one?

hawaiidude

lol no it is correct

Integral

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

y=4x

Not real exicting.

ExtravagantDreams

Yeah, that's what I was thinking.

ExtravagantDreams

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^&lambda;tcos(&mu;t) + C₂e^&lambda;tsin(&mu;t)

for complex roots in the form;

r = &lambda; ± i&mu;

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 &plusmn; &radic;(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