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December 24th, 2014, 05:06 AM   #1
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differential equation

solve the equation:

dy/dx +sin(dy/dx)=x and y=0 when x=0.

this is the last question on the problem sheet and nothing else like it nor have we been taught how to approach something like this. I cant see how to start this at all. any ideas?
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December 24th, 2014, 08:24 AM   #2
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Can you check for typing errors or post a picture?
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December 25th, 2014, 12:13 AM   #3
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its question 10 on the attachment.
may i ask why you think there is a typo?
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December 25th, 2014, 06:45 AM   #4
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Hello, fromage!

Quote:
$\text{Solve the equation: }\;\dfrac{dy}{dx} + \sin\left(\dfrac{dy}{dx}\right) \:=\:x,\quad \text{with }y(0) = 0.$

We are concerned because the equation
$\quad$has the form: $\:u + \sin u \,=\,x$

This is a transcendental equation in which a variable is both
$\quad$ 'inside' and 'outside' of a transcendental function.

It cannot be solved by conventional means.
$\quad$The solution can only be approximated.

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December 25th, 2014, 06:46 AM   #5
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Because at first sight it looks likely to be impossible to solve!
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December 27th, 2014, 12:15 AM   #6
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ok guys my professor got back to me and he said it could be solved parametrically in terms of dy/dx.
ie. let u=dy/dx

x=u+sinu
(differentiating gives)=> 1=du/dx(1+cosu)
=> 1=u*du/dy(1+cosu)
=> y=integral of (u(1+cosu))du
(after doing integral
and putting in condition) => y= 1/2*(dy/dx)^2+(dy/dx)*sin(dy/dx)+cos(dy/dx)-1

so here's my new question- is this useful? (i can see it answers the question but does it give us more information; to me it seems it does but i don't know if having both x and y in terms of dy/dx can help us more)

and if it is useful (and i mean no offence here at all to any of you, i'm just curious as to the reasoning behind this method) why didn't you guys see this?
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December 29th, 2014, 11:25 PM   #7
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bump?hello?
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December 30th, 2014, 12:33 PM   #8
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Mm... I wouldn't call that solving, but arranging the equation to another form. It is not clear, which one is more useful and it will strongly depend on the situation.

If one really needs to get numbers out, then I'd use the first form of the equation. It is much more simpler and needs less computations. I'd also think it is more accurate, if you need the $\displaystyle (x,y)$ pairs because you can solve the derivative just by giving the value of $\displaystyle x$. As for the second form, I can't find use for it quite fast. Maybe you or your professor can?

Also, if you really are interested, the solution ($\displaystyle y'$) of first form of the equation can be written as an infinite serie in terms of Bessel functions. This serie can further be integrated and you have a solution in some form.
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