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Jack Rod

2:55 AM

I am really confuse in it sir if u can take out some time for this

@robjohn my book uses two methods two solve $(\frac{1}{D-1}-\frac{1}{D-2})e^{5x}$

one using linear differential equation of first order

and second is partial fraction

and I have problem in understanding both pf them

Jack Rod

3:12 AM

ping me whenever u get a time

Jack Rod

3:27 AM

@robjohn

robjohn

4:04 AM

@JackRod There are no real operators $\frac1{D-1}$ and $\frac1{D-2}$, so the difference doesn't really offer any help. However, we can invert operator $D-a$.

Suppose we have $(D-a)y=f(x)$. We can use integrating factors:
$$
\begin{align}
gf
&=\overbrace{gDy}^{gy'}+\overbrace{-gay\vphantom{D}}^{g'y}\\
&=D(gy)
\end{align}
$$
as long as $Dg=-ag$

so we can use $g=e^{-ax}$

Then $$\overbrace{e^{ax}\vphantom{f}}^{g^{-1}}\int\overbrace{e^{-ax}f}^{gf}\,\mathrm{d}x=y$$

Jack Rod

3 mins ago, by robjohn

Suppose we have $(D-a)y=f(x)$. We can use integrating factors:
$$
\begin{align}
gf
&=\overbrace{gDy}^{gy'}+\overbrace{-gay\vphantom{D}}^{g'y}\\
&=D(gy)
\end{align}
$$
as long as $Dg=-ag$

robjohn

surely you can solve $Dg=-ag$

$\frac{Dg}{g}=-a$

$\log(g)+C=-ax$

Jack Rod

Sir I am bit lost you are solving it using linear differential equation like dy/dx+y=Q

robjohn

That would be $(D+1)y=Q$

Jack Rod

sorry (D-1)y=Q

i.f is $e^{-x}$

robjohn

4:12 AM

So if we have $gf=D(gy)$ we can solve for $y$

Jack Rod

ok sir if it have three roots

robjohn

Please try to understand this first.

I am trying to show you how to invert $D-a$

Jack Rod

robjohn

Suppose we have $(D-a)y=f(x)$.

Jack Rod

robjohn

4:15 AM

multiply by $g$: $gDy-gay=gf$

If we have that $Dg=-ga$, then $gf=gDy-gay=gDy+yDg=D(gy)$

integrate: $\int gf\,\mathrm{d}x=gy$

Jack Rod

yes sir

@robjohn

Jack Rod

4:36 AM

are u there?

I understood what u are trying to explain above

Jack Rod

5:08 AM

@robjohn

Jack Rod

5:40 AM

$(D^2-a^2)=cosh ax $

how we gonna do it

2 hours later…

Jack Rod

7:19 AM

@robjohn hi

robjohn

7:39 AM

@JackRod that is $(D-a)(D+a)y=\cosh(ax)$, so you invert $D+a$ and $D-a$ as we did above.

Invert $D-a$ with $e^{ax}\int e^{-ax}f(x)\,\mathrm{d}x$

Invert $D+a$ with $e^{-ax}\int e^{ax}f(x)\,\mathrm{d}x$

So $(D-a)y=e^{-ax}\int e^{ax}\cosh(ax)\,\mathrm{d}x=\frac12e^{-ax}\int \left(e^{2ax}+1\right)\mathrm{d}x=\frac1{4a}e^{ax}+\frac x2e^{-ax}+Ce^{-ax}$

Jack Rod

ok sir got it

@robjohn $d^3y/dx^3+y=x^3+sin3x$

it is coming with imaginary roots

robjohn

you can integrate complex exponents

Jack Rod

I wrote it as $(D^3+1)=y+sin3x$

robjohn

$\left(D^3+1\right)y=x^3+\sin(3x)$

Jack Rod

then D has root slike -1,$w^2,w$

robjohn

7:51 AM

so you have three operators to invert

Jack Rod

yes

robjohn

two with complex exponent multipliers

But you handle it the same way, inverting $D-a_1$ then $D-a_2$ then $D-a_3$

The integrals are indefinite, so remember the constants of integration

it might simplify things to solve $\left(D^3+1\right)y=x^3$ and $\left(D^3+1\right)y=\sin(3x)$ separately and then add the solutions.

at least you will then be dealing with smaller functions

Jack Rod

sir by doing we will get exact differential of this?

robjohn

excat?

Jack Rod

what I am saying is I will get something like $y=c_1e^{-x}+c_2e^{k}+c_3e^{l}$

so I will solve for real integration that is

$(D^3+1)y=x^3$

is it same like we solve linear differential equation of first order?

@robjohn

robjohn

8:13 AM

@JackRod you won't be getting those for the $x^3$ integrals, but you will for the $\sin(3x)$ integrals.

@JackRod There will be a complex exponentials in the solutions of that

@JackRod not sure exactly what you mean here.

Jack Rod

$\frac{1}{D^3+1}{e^x}+\frac{1}{D^3+1}\sin3x$

my book simply wrote it as $\frac{1}{2}e^x+\frac{1}{-3^2.D+1}\sin3x$

I do not how they wrote it like this?

robjohn

I don't either, as the second summand does not make sense.

Jack Rod

may be using binomial

4 mins ago, by Jack Rod

$\frac{1}{D^3+1}{e^x}+\frac{1}{D^3+1}\sin3x$

sir how will u solve this part?

robjohn

As I said, those don't really make sense, but what they are saying is what I said above:

28 mins ago, by robjohn

it might simplify things to solve $\left(D^3+1\right)y=x^3$ and $\left(D^3+1\right)y=\sin(3x)$ separately and then add the solutions.

solve each of those by inverting $D+1$, $D+\omega$ and $D+\omega^2$

Jack Rod

you mean $(D+1)y=x^3$

robjohn

8:29 AM

no, I meant $\left(D^3+1\right)y=x^3$

Jack Rod

I do not know how to solve it

it is 3rd order differential equation

robjohn

41 mins ago, by Jack Rod

@robjohn $d^3y/dx^3+y=x^3+sin3x$

This is the equation you first stated and it is $\left(D^3+1\right)y=x^3+\sin(3x)$

I don't know where the $e^x$ came into the solution

Jack Rod

9:04 AM

leave this question sir

@robjohn can we move to next one?

Jack Rod

9:20 AM

@robjohn can u expand it for me

\begin{align}
(D-2)^{-1} e^{5x}
&=-2(1-D/2)^{-1} e^{5x}\\
&=-2\left(1+\frac12 D+\frac14 D^2+\cdots\right )e^{5x}\\
&=-2(1+5/2+(5/2)^2+\cdots)e^{5x}\\
&=-2(1-5/2)^{-1} e^{5x}\\
&=\frac43 e^{5x}
\end{align}

is it right?

it was said by @semiclassical

and when I asked him about why he choose D=5 he said

we haven't used D=5 as such. what we've used is that $De^{5x}=5e^{5x}\implies D^2 e^{5x}=5^2 e^{5x}$ and so forth
so $D\neq 5$ when acting on arbitrary functions, but $D$ does behave like "multiplication by 5" when acting on $e^{5x}$
this is easy enough to justify at the level of $(D-2)e^{5x}=3e^{5x}$
the 'operator series' method is where things get dicey in terms of rigour

@robjohn I cannot understand what he meant by this?

8 hours later…

robjohn

5:40 PM

This might be able to be worked up into a rigorous argument, but, unless you've done a bit of work showing why the series converges, using series with differential operators is often used to guess an answer, and then verify that that is an answer.

However, the answer you get is wrong since the answer is $\frac13e^{5x}$ (just apply $D-2$ to it).

robjohn

6:13 PM

My guess is that that answer doesn't work because that series just doesn't converge. $1+\frac52+\left(\frac52\right)^2+\left(\frac52\right)^3+\dots$ diverges.

robjohn

6:51 PM

It should really be handled by inverting $D-\sqrt2$ and $D+\sqrt2$.

Using $e^{\sqrt2\,x}\int e^{-\sqrt2\,x}f(x)\,\mathrm{d}x$ and $e^{-\sqrt2\,x}\int e^{\sqrt2\,x}f(x)\,\mathrm{d}x$ respectively

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