Solve the integro-differential equation
$$\int\limits_0^xy(u)\,\mathrm du-y'(x)=x$$

Question

Solve the integro-differential equation
$$\int\limits_0^xy(u)\,\mathrm du-y'(x)=x$$

unklerhaukus · Answer

i've got this,

unklerhaukus · Answer

but 
the answer in the back of the book says 
$$=y_0\cosh x-\cosh x+1\
= (y_0-1)\cosh x+1$$

zzr0ck3r · Answer

@zzr0ck3r googles integro-differential  to see if it is taken for a band name. NOPE!

kainui · Answer

Maybe this is the problem, but I don't think it should matter so I'll keep looking.
$$Y=(y_0-X)\frac{p}{p^2-1} = y_0 \frac{p}{p^2-1} +\frac{1}{1-p^2}$$

kainui · Answer

Oh nevermind, I think my correction is wrong, I was thinking $X=\frac{1}{p}$ but really it's $\frac{1}{p^2}$ isn't it. I guess that's lame, and I suppose you're doing the convolution to avoid partial fractions so I don't blame you.

kainui · Answer

All of your work looks completely correct to me. Maybe it can be rearranged from one into the other ugh lol.

unklerhaukus · Answer

sinh = cosh 
?

kainui · Answer

Or your book is wrong. We can try checking to see if the answers satisfy the integro-differential equation.

kainui · Answer

The book answer is right, your answer is wrong in general but true with $y_0=0$ so that might give some clues.

ganeshie8 · Answer

By a different method im getting
y(x) = C*coshx + 1

unklerhaukus · Answer

hmmm

kainui · Answer

I am not sure but I don't think the inverse laplace transform was done correctly here:

$$\mathcal{L}^{-1}\{y_0-X\} = y_0-x$$

ganeshie8 · Answer

I'll need to review laplace transofrms to understand your method, but your y(x) is clearly wrong as it doesn't satisfy the given equation

unklerhaukus · Answer

what is the inverse Laplace of a constant?

kainui · Answer

Dirac delta? Not sure

unklerhaukus · Answer

oh, yeah the delta thingo

kainui · Answer

seems like that could plausibly fix it, not gonna work it out though unless you say it doesn't work haha

unklerhaukus · Answer

how could partial fractions be used instead?

kainui · Answer

Well if you expand X and avoid the convolution, then you end up with:

$$\frac{1}{p^2} \frac{p}{p^2-1} = \frac{1}{p(p^2-1)} = -\frac{1}{p}+\frac{1}{2} \frac{1}{p+1}+\frac{1}{2} \frac{1}{p-1} $$

then just transform these terms individually.

kainui · Answer

Looks like these will inverse transform into some exponentials, which is fine cause we can make cosh and sinh out of exponentials.

unklerhaukus · Answer

What method did you use @ganeshie8 ?

ganeshie8 · Answer

Differentiate through out, solve the second order ode, then reduce the constants...

ganeshie8 · Answer

The solutions of
$$\int\limits_0^xy(u)\,\mathrm du-y'(x)=x\tag{1}$$
are a subset of the solutions to
$$y - y'' = 1\tag{2}$$

Not hard to see that $y = 1$ is a particular solution to $(2)$
The null solution is $c_1e^x + c_2e^{-x}$.
The complete solution to $(2)$ is $y = 1+c_1e^x + c_2e^{-x}$

Plugging this in $(1)$ gives $c_1=c_2$, so the general solution to $(1)$ has to be $y = c(e^x + e^{-x})+1$, which can be rearranged as $y = C* \cosh x + 1$

unklerhaukus · Answer

I can now get your method to work @ganeshie8 , it's not hard to solve 2nd order DE.
Thank you for providing it.

However I would still like to fix my Laplace convolution method

unklerhaukus · Answer

I think you are right about my error @Kainui 
$$\mathcal{L}^{-1}\{y_0-X\} \neq y_0-x$$
but i'm not really sure how to fix this

kainui · Answer

Alright I'll pick it up right where you left it off and give you some stuff to finish it on your own I think cause it should boil down to just distributing mostly the same work you have already done.

$$Y= (y_0-X)\frac{p}{p^2-1}$$$$y=\mathcal{L}^{-1}[\mathcal{L}\{y_0\delta(x) -x\}\mathcal{L}\{\cosh(x)\}] = (y_0 \delta(x)-x) \star \cosh(x) $$

Actually, if you don't know how to derive the convolution, I think it's definitely worth your time cause it's pretty ubiquitous, it's a way of sidestepping multiplying two series together and looking at only their coefficients and is sorta a meeting point of linear algebra and group theory.

Ok so on with the calculation...

$$y =\int_0^x [y_0 \delta(x-u) - (x-u)] \cosh(u)du$$
One of these integrals you have already solved so I'll just give you what you need:
$$\int_0^x \delta(x-u) f(u)du = f(x)$$

You can think of $\delta(0)=\infty$ and all other values $\delta(t)=0$. It's essentially the identity matrix, and is the identity element of the group of functions with the convolution as the binary operation of this abelian group... Too much information, w/e throw it away lol. That should be enough to solve it now:


$$y =y_0 \cosh(x)+\int_0^x - (x-u) \cosh(u)du$$

kainui · Answer

Yeah continuing this on, I am getting the exact answer as your book now. Awesome.

kainui · Answer

Just to sorta show the forward way, since x=0 is the only value that $\delta(x)$ is nonzero, it "pulls out" that value:
$$\mathcal{L}\{\delta\}(p)=\int_0^\infty \delta(x) e^{-px} dx = e^{-p0}= 1$$

unklerhaukus · Answer

$$\boxed{
\newcommand \dd	 [1] {\,\mathrm d#1}					%  infinitesimal
\newcommand \intl[4] {\int\limits_{#1}^{#2}{#3}{\dd #4}}		%  integral_a^b{f(x)}dx 
\newcommand \Lap [1] {\operatorname{\mathcal L}\left\{#1\right\}}	%  Laplace transform (p)
\newcommand \Lin [1] {\operatorname{\mathcal L}^{-1}\left\{#1\right\}}	%  inverse Laplace transform
\newcommand \Ev  [3] {\left.{#1}\right|_{#2}^{#3}}			%  evaluation limits
\newcommand \ps  [1] {\left(#1\right)}					%  dynamic parentheses
}$$

$$
\begin{align}
    \intl0x{y(u)}u-y'(x)
    	&= x\
    \Lap{1\star y}-\Lap{y'}
    	&= X\
    \frac1p\cdot Y	- \ps{pY-y_0}
    	&=\
    \ps{\frac1p-p}Y
    	&= X-y_0\
    Y	&= \ps{X-y_0}{\frac1{\frac1p-p}}\
    	&= \ps{y_0-X}{\frac p{p^2-1}}\
    y(x)&= \ps{y_0\delta(x)-x}\star\cosh x\
    	&= \intl0x{(y_0\delta(x-u)-(x-u))\cosh u}u\
    	&= y_0\intl0x{\delta(x-u)\cosh u}u-x\intl0x{\cosh u}u
		+ \intl0x{u\cosh u}u\
    	&= y_0\Ev{\cosh u}{}x-x\Ev{\sinh u}0x
		+ \Ev{u\sinh u}0x-\intl0x{\sinh u}u\
    	&= y_0\cosh x-x\sinh x
		+ x\sinh x-\Ev{\cosh u}0x\
    	&= y_0\cosh x - \cosh x + 1\
    	&= \ps{y_0-1}\cosh x+1
\end{align}
$$

unklerhaukus · Answer

We have
$$y(x) = (y_0-1)\cosh x+1$$
Taking the derivative 
$$y'(x)=(y_0-1)\sinh x$$
and integral: 
$$\intl0x{y(u)}u= (y_0-1)\intl0x{\cosh u+1}u\
\qquad\qquad = (y_0-1)\sinh x+x$$

Plugging these into the IDE
$$\int\limits_0^xy(u)\,\mathrm du-y'(x)=x\
$$$$LHS
= \Big[(y_0-1)\sinh x+x\Big]-\Big[ (y_0-1)\sinh x\Big] \
\qquad = x\
\qquad = RHS$$

irishboy123 · Answer

vanilla way, starting at second line

$\dfrac{1}{p} Y - p Y + y_o = \dfrac{1}{p^2}$

$Y = \dfrac{p}{p^2 -1^2} y_o - \dfrac{1}{p} \dfrac{1}{p^2 - 1^2}$


$ =  \mathcal{L} \{ y_o \cosh x\} - \mathcal{L} \{1\}  \mathcal{L} \{\sinh x\}$

Second bit is convolution

$y =  y_o \cosh x - \int\limits_0^x \sinh \tau d \tau$

$y =  y_o \cosh x -  \left. \cosh \tau \right|_0^x$

unklerhaukus · Answer

That's much better. @IrishBoy123