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18 / 4 / 10 / ?

File: Linear-Differential-Equation.png ( 667.17 KB , 2000x2068 , 1655761703189.png )

Image
  No. 207
>>626
Post diff eqs and their solutions.
>>
  No. 208
Linear differential equation of the first order:

y′+Py=Q\qquad \boxed{y' + Py = Q}


General solution:
Multiply both sides with
e∫Pdxe^{\int P\mathrm{d}x}


y′e∫Pdx+Pye∫Pdx=Qe∫Pdx\qquad y'e^{\int P\mathrm{d}x} + Pye^{\int P\mathrm{d}x} = Qe^{\int P\mathrm{d}x}


Collapse left side
(fg)′=f′g+fg′(fg)' = f'g + fg'


(ye∫Pdx)′=Qe∫Pdx\qquad \left(ye^{\int P\mathrm{d}x}\right)' = Qe^{\int P\mathrm{d}x}


Integrate both sides

ye∫Pdx=C+∫Qe∫Pdxdx\qquad ye^{\int P\mathrm{d}x} = C + \int Qe^{\int P\mathrm{d}x}\mathrm{d}x


Multiply both sides with
e−∫Pdxe^{-\int P\mathrm{d}x}


y=e−∫Pdx(C+∫Qe∫Pdxdx)\qquad \boxed{y = e^{-\int P\mathrm{d}x}\left(C + \int Qe^{\int P\mathrm{d}x}\mathrm{d}x \right)}



Example:

y′+2xy=4x\qquad \boxed{y' + 2xy = 4x}


Solution:
Since
P=2x,Q=4xP = 2x,\enspace Q=4x
:
y=e−x2(C+∫4xex2dx)=e−x2(C+2ex2)=Ce−x2+2\qquad y = e^{-x^2}\left(C + \int 4xe^{x^2}\mathrm{d}x\right) = e^{-x^2}(C + 2e^{x^2}) = \boxed{Ce^{-x^2} + 2}
>>
  No. 209 >>1005
Bernoulli's differential equation:

y′+Py=Qyα\qquad \boxed{y' + Py = Qy^\alpha}


General solution:

Sbustitute
y=zry=z^r


rzr−1z′=Pzr=Qzrα\qquad rz^{r-1}z' = Pz^r = Qz^{r\alpha}


Divide both sides with
rzr−1 rz^{r-1}


z′+Prz=Qrzrα−r+1\qquad z' + \frac{P}{r} z = \frac{Q}{r}z^{r\alpha - r +1}


Setting
rr
such that
rα−r+1=0r\alpha - r + 1 = 0
(i.e.
r=11−αr = \frac{1}{1-\alpha}
) Bernoulli's equation becomes a linear differential equation of the first order that can be solved for
zz
.

z=e−(1−α)∫Pdx(C+(1−α)∫Qe(1−α)∫Pdxdx)\qquad z = e^{-(1-\alpha)\int P\mathrm{d}x}\left(C + (1-\alpha)\int Qe^{(1-\alpha)\int P\mathrm{d}x}\mathrm{d}x\right)


Substituting back
y=zr,y=z11−α,z=y1−αy = z^r,\quad y=z^{\frac{1}{1-\alpha}},\quad z = y^{1-\alpha}
:

y=(e−(1−α)∫Pdx(C+(1−α)∫Qe(1−α)∫Pdxdx))11−α\qquad \boxed{y = \left({e^{-(1-\alpha)\int P\mathrm{d}x}\left(C + (1-\alpha)\int Qe^{(1-\alpha)\int P\mathrm{d}x}\mathrm{d}x\right) }\right)^{\frac{1}{1-\alpha}}}



Example:

y′−y=xy5\qquad y' - y = xy^5


Solution:

Substituting
y=z−1/4,y′=−14z−5/4z′y = z^{-1/4},\quad y' = -\frac{1}{4}z^{-5/4}z'
:
−14z′−z=x\qquad -\frac{1}{4} z' - z = x


Solving this linear quation for
zz
and substituting
z=y−4z = y^{-4}
should yield

y−4+x−14=Ce−4x\qquad y^{-4} + x - \frac{1}{4} = Ce^{-4x}


\qquad
>>
  No. 210 >>212 >>215 >>646
Riccati differential equation equation

y′=Py2+Qy+R\qquad \boxed{y' = Py^2 + Qy + R}


General solution
If
P≡0P\equiv 0
reduces to a linear differential equation and if
Q≡0Q\equiv 0
it reduces to Bernoulli's equation.

Solving Riccati equation using simple integration is not possible unless one at least one particular solution
y1y_1
is known, in which case a general solution is obtained by substituting
y=y1+1zy = y_1 + \frac{1}{z}

y1′−z′z2=P(y12+2y1z+1z2)+Q(y1+1z)+R\qquad y_1' - \frac{z'}{z_2} = P\left(y_1^2 + \frac{2y_1}{z} + \frac{1}{z^2}\right) + Q\left(y_1 + \frac{1}{z}\right) + R


Because
y1y_1
is known to be a solution,
y1′=Py12+Qy1+Ry_1' = Py_1^2 + Qy_1 + R
will cancel resulting in:

z′+(2Py1+Q)z+P=0\qquad z' + (2Py_1 + Q)z + P = 0


Solving this linear equation results in a solution of the form
z=z(x,C)z=z(x,C)
, so the general solution is:

y=y1+1z(x,C)\qquad y = y_1 + \frac{1}{z(x, C)}


A particular solution
y1y_1
can be obtained from the general solution by letting
C→±∞C\rightarrow\pm\infty
.

Example

y′=y2−xy+1\qquad y' = y^2 -xy + 1


Assuming one particular solution is in the form of
y1=ax+by_1 = ax + b
.

Solution
Substituting
y=ax+by = ax+ b


a=a2x+2abx+b2−ax2−bx+1\qquad a = a^2x + 2abx + b^2 - ax^2 - bx + 1

a2−a=0,2ab−b=0,a−1=0\qquad a^2 - a = 0,\quad 2ab - b=0,\quad a-1 =0

a=1,b=0\qquad a=1, b= 0


Thus
y1=xy_1 = x
is a particular solution.
Now substituting
y=x+1zy = x + \frac{1}z{}


1−z′z2=x2+2xz+1z2−x2−xz+1,z′+xz=−1\qquad 1 - \frac{z'}{z^2} = x^2 + \frac{2x}{z} + \frac{1}{z^2} - x^2 -\frac{x}{z} + 1,\quad z'+ xz = -1


Solving
z′+xz=−1z'+xz = -1
with the formula
z=e−∫xdx(C−∫e−∫xdxdx)z = e^{-\int x\mathrm{d}x}\left(C - \int e^{-\int x\mathrm{d}x}\mathrm{d}x\right)

z=e−x22(C−∫ex22dx)\qquad z = e^{-\frac{x^2}{2}}\left(C - \int e^\frac{x^2}{2}\mathrm{d}x\right)


Thus the general solution to the differential equation is:

y=x+ex22C−∫ex22dx\qquad y = x + \frac{e^{\frac{x^2}{2}}}{C - \int e^\frac{x^2}{2}\mathrm{d}x}


The integral
∫ex2/2dx\int e^{x^2/2}\mathrm{d}x
is the Gaussian integral which of course is unsolvable.
>>
  No. 212 >>213 >>646
>>210
Can you elaborate more on the unsolvability of the Gaussian integral? Can we define a function to be the solution of it or is it just not possible?
>>
  No. 213 >>646

File: Plot-of-the-Gauss-error-function-erfx.png ( 12.74 KB , 722x301 , 1656012118622.png )

Image
>>212
Well, there is certainly an antiderivative function for
f(x)=e−x2f(x)=e^{-x^2}
and you can plot it (pic rel), it just can't be written in terms of elementary functions. Elementary function is any finite composition of addition, multiplication, polynomials (
1,x2,x3,…1,x^2,x^3,\dots
), trigonometric and inverse trigonometric functions (
sin⁡(x),cos⁡(x),tan⁡(x),…\sin(x), \cos(x), \tan(x),\dots
,
arcsin⁡(x),arccos⁡(x),arctan⁡(x),…\arcsin(x),\arccos(x),\arctan(x),\dots
), powers (
xax^a
), exponentials (
exe^x
) and logarithms (
log⁡(x)\log(x)
) i.e. everything you learn in high school.

There are plenty of functions whose antiderivatives can't be written in terms of elementary functions.

>Can we define a function to be the solution
Yes. Since you can't write the solution
e−x2e^{-x^2}
in terms of elementary functions, the integral called "unsolvable", but you can define a new function to say "this function is the solution to this integral". And in fact, that's the error function.

erf(x)=2π∫0xe−x2dx\qquad \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^xe^{-x^2}\mathrm{d}x


There are many other functions defined this way (e.g. the gamma function
Γ(x)=∫0∞xn−1e−xdx\Gamma(x) = \int_0^{\infty}x^{n-1}e^{-x}\mathrm dx
) and they're used daily, but they are not considered elementary functions in the usual sense and neither Bessel functions
Jα(z),Yα(z)J_\alpha(z),Y_\alpha(z)
, Lambert
W(z)W(z)
function, sine and cosine integral functions
Si(x),Ci(x)\mathrm{Si}(x), \mathrm{Ci}(x)
, Fresnel integral functions
S(x),C(x)S(x), C(x)
etc.
>>
  No. 215 >>218
>>210
>Solving Riccati equation using simple integration is not possible unless one at least one particular solution is known
I think this is somewhat misleading. There are plenty of techniques available to handle nonlinear differential equations.
I think a helpful comment here is that it is often useful to use a linearising substitution like
y=ψ′/ψy=\psi'/\psi
which in some cases will reduce a Ricatti equation to a linear equation for
ψ\psi
.
Without loss of generality, you can always reduce Ricatti equations to the form where the coefficient of
y2y^2
is 1, (substitute for
y↦Pyy\mapsto Py
) and then you can apply the substitution above. This linearises the equation.
This is an essential idea to know when handling the Ricatti equation and nonlinear differential equations in general.
>>
  No. 218
>>215
typo, meant to say
y=−ψ′/ψy=-\psi'/\psi
>>
  No. 235 >>240
So how do you learn solutions to DEs? Do you just apply them over and over again until you've got a feeling for them? I can always follow them just like I can follow proofs, but being able to freely write them takes an unusual amount of effort. Much more than just memorizing a few facts. Do you guys think flashcards with random parts blurred out could do it?
>>
  No. 240 >>243
>>235
I think it depends. Generally I think the best way to learn maths is to do it until you develop a gut feel for it. Linear differential equations are normally not too hard to develop a feel for, because you'll run into many of them in standard practical applications. Sometimes you'll learn standard ideas for bringing equations from one form into another.
I don't think flashcards would be useful for this.
Nonlinear differential equations are much harder. Sometimes they're unsolvable. When they are solvable, there is no general method which will work for other nonlinear equations of a different type (which you can always do with linear equations, by contrast). This means that the field is replete with clever trickery which you either know or you don't. For example, you either know that the viscous Burger's equation can be handled by the Cole-Hopf transformation, or you can go and beat your head against a wall until you figure it out on your own. Once you know that, it might help you with other nonlinear differential equations (e.g. the Ricatti equation) but it will be utterly useless for others (e.g. the inviscid Burger's equation, or Liouville's equation, although a similar substitution works for the latter, or many other equations which will not be solvable by any trick you can think of).
>>
  No. 243 >>244
>>240
Thanks for your reply. I was only thinking of flashcards since I had massive success with Anki in the past, in both language learning as well as memorization-heavy subjects, like biology and history. After playing around with it in math for a while, I quickly realized that cards that take a long time to answer or have a billion clozes aren't that great/don't work. Maybe cards querying the basic idea behind a solution to a DE could work? I just want to prevent forgetting things once I've passed the exam.
There's also this paper I found http://uweb.cas.usf.edu/~drohrer/pdfs/Rohrer&Taylor2006ACP.pdf (what about a card deck that just reminds you to do certain exercises or review them months after having done them?)
>>
  No. 244 >>251
>>243
>Maybe cards querying the basic idea behind a solution to a DE could work?
I think this is a good idea. You should of course learn in whatever way suits you, rather than letting other people tell you how to learn, but in my opinion it is much more important to understand the general principle behind what you are doing with a certain technique. This gives you deeper insight into what is really going on, and is also easier to remember. You might forget the exact expression/method you want, but you can replicate it on demand so long as you can understand the idea, and you can also bring the same insight to bear on other equations.
>>
  No. 251
>>244
Of course, I will make sure to do this, in fact, trying to memorize something that you don't understand will not work. Really, all Anki does, for me, is have me review the notes I made when reading textbooks, with the added benefit of scheduling what to review at what time; it's not really rote memorization. Every now and then, because I also put screenshots of the pages I extracted the information from on the back of the card as an extra (I won't try exactly recalling that, obviously), I realize something new/connections to newer cards in old cards/information.
And its reviews are perfect for me since they're completely devoid of any context. You're actively trying to recall a concept just as if a stranger randomly approached you on the street and asked you (this is really important. I had made the mistake of telling the program "good" when I couldn't recall it but the content, once shown to me, was familiar, before, which significantly slowed down my learning). As such, I believe active recall would rather aid the ability to replicate it on demand as it has you replicate/recall it, and that just when you're about to forget it.

I'll try it out more seriously for math the next few months and report back.
>>
  No. 585
Here's a fun one:
∂tu+(u⋅\nablda)u−νΔu+\gradp=f\partial_t u+(u\cdot\nablda)u-\nu\Delta u+\grad p=f

div  u=0\mathbf{div}\;u=0

I'll leave it to the next person to post the solution.
>>
  No. 605
nice
>>
  No. 626

File: FTDC9FGXwAEZ211.jpg ( 297.09 KB , 2048x1616 , 1705954986668.jpg )

Image
>>207
how did i not see this lol i have a wide collection

Consider
y′′−2y=ln⁡(x)y''-2y=\ln(x)
.

Let's first solve for the complementary solution, which is the general solution to the complementary equation
y′′−2y=0y''-2y=0
.

The characteristic polynomial for the complementary equation is
λ2−2=0\lambda^2-2=0
, so
λ=±2\lambda=\pm \sqrt{2}
. Hence, the complementary solution is
yc(x)=Ay1(x)+By2(x)=Ae−x2+Bex2y_c(x)=Ay_1(x)+By_2(x)=Ae^{-x\sqrt{2}}+Be^{x\sqrt2}
for some constants
AA
and
BB
. Now let's find the particular solution using variation of parameters. Using the two basis solutions of the complementary solution,
y1(x)y_1(x)
and
y2(x)y_2(x)
, we construct the Wronskian and get
W(e−x2, ex2)(x)=∣e−x2ex2(e−x2)′(ex2)′∣=22W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x) = \begin{vmatrix} e^{-x\sqrt{2}} & e^{x\sqrt{2}}\\ \left(e^{-x\sqrt{2}}\right)' & \left(e^{x\sqrt{2}}\right)' \end{vmatrix} = 2\sqrt2


Hence, for the particular solution
yp(x)=u1(x)y1(x)+u2(x)y2(x)y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)
, we have,
u1(x)=−∫ln⁡(x)y2(x) dxW(e−x2, ex2)(x)=−∫ln⁡(x)ex2 dx22=14Ei⁡(x2)−14ex2ln⁡(x) u_1(x) = - \int\frac{\ln(x)y_2(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = -\int\frac{\ln(x)e^{x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)

u2(x)=∫ln⁡(x)y1(x) dxW(e−x2, ex2)(x)=∫ln⁡(x)e−x2 dx22=14Ei⁡(−x2)−14e−x2ln⁡(x) u_2(x) = \int\frac{\ln(x)y_1(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = \int\frac{\ln(x)e^{-x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)


The particular solution then is nothing but
yp(x)=u1(x)y1(x)+u2(x)y2(x)=(14Ei⁡(x2)−14ex2ln⁡(x))e−x2+(14Ei⁡(−x2)−14e−x2ln⁡(x))ex2=14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) y_p(x) = u_1(x)y_1(x)+u_2(x)y_2(x) = \left(\frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)\right) e^{-x\sqrt2} + \left(\frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)\right) e^{x\sqrt2}\\ = \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)

Thus the complete general solution is given by
y(x)=yc(x)+yp(x)=Ae−x2+Bex2+14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) \boxed{y(x) = y_c(x) + y_p(x) = Ae^{-x\sqrt{2}}+Be^{x\sqrt2} + \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)}
>>
  No. 646
>>210
>>212
>>213

Honestly, I never understood why we defined the notion of an "elementary" function the way we did. An elementary function from what I've seen is completely restricted to polynomials/rational powers, exponentials, and logarithms (trig/hyperbolic trig and their inverses are expressible in exponentials/logs), which is extremely limited.
>>
  No. 655

File: GFv9PYsakAA1vJW.jpg ( 1.28 MB , 1400x1900 , 1708564466975.jpg )

Image
We have
y′=(1+x) y+xy2y'=(1+x)\,y+xy^2
.

Rewriting this gives
y′−(1+x) y=xy2y'-(1+x)\,y=xy^{2}
. This is a standard Bernoulli diffeq form
y′+a(x) y=b(x) yαy'+a(x)\,y=b(x)\,y^{\alpha}
, which is solved by performing the substitution
ϕ=y1−α\phi=y^{1-\alpha}
. Using this substitution, which for our specific case is
ϕ=y1−2=y−1\phi = y^{1-2} = y^{-1}
, we have
−y′y2+x+1y=−x⟺ϕ′+(x+1) ϕ=−x\displaystyle -\frac{y'}{y^2} +\frac{x+1}{y} = -x \Longleftrightarrow \phi' + (x+1)\,\phi = -x
.

To make this substitution more rigorous (since it is less obvious), differentials can be used. Also, to account for this division and ensure we miss no solutions, we make sure to check when the denominators are equal to 0, which is
y=0y=0
. Since plugging in
y=0y=0
satisfies the diffeq, this is a constant solution. Then, using the integrating factor
exp⁡(∫x+1 dx)=ex22+x\exp\left(\displaystyle\int x+1\text{ d}x\right) = e^{\frac{x^2}{2}+x}
, we get
ϕ′ex22+x+(ex22+x)′ϕ=−xex22+x∫(ϕ ex22+x)′dx=∫−xex22+x dxϕ ex22+x=π2eerfi⁡(x+12)−ex22+x+Cϕ=π2eerfi⁡(x+12)−ex22+x+Cex22+x \begin{align*} \phi'e^{\frac{x^2}{2}+x} + \left(e^{\frac{x^2}{2}+x}\right)'\phi &= -xe^{\frac{x^2}{2}+x}\\ \int\left(\phi \,e^{\frac{x^2}{2}+x}\right)'\text{d}x &= \int-xe^{\frac{x^2}{2}+x}\text{ d}x\\ \phi \, e^{\frac{x^2}{2}+x} &= \sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C\\ \phi &= \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} \end{align*}

Undoing our substitution, we have, for the positive Dawson Integral
D+(x)=π2e−x2erfi⁡(x)D_+(x) = \frac{\sqrt{\pi}}2e^{-x^2}\operatorname{erfi}(x)
that
ϕ=y−1=π2eerfi⁡(x+12)−ex22+x+Cex22+x=2 D+(x+12)−1+Ce−x22−x\displaystyle \phi = y^{-1} = \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} = \sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}

  ⟹  y(x)=0,12 D+(x+12)−1+Ce−x22−x\displaystyle\implies \boxed{y(x)=0, \qquad \frac{1}{\sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}}}
>>
  No. 1005

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