
Differential Equations Ordinary and Partial Differential Equations Math Forum 
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December 24th, 2014, 06:06 AM  #1 
Member Joined: Oct 2013 Posts: 36 Thanks: 0  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? 
December 24th, 2014, 09:24 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra 
Can you check for typing errors or post a picture?

December 25th, 2014, 01:13 AM  #3 
Member Joined: Oct 2013 Posts: 36 Thanks: 0 
its question 10 on the attachment. may i ask why you think there is a typo? 
December 25th, 2014, 07:45 AM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, fromage! Quote:
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.  
December 25th, 2014, 07:46 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra 
Because at first sight it looks likely to be impossible to solve!

December 27th, 2014, 01:15 AM  #6 
Member Joined: Oct 2013 Posts: 36 Thanks: 0 
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? 
December 30th, 2014, 12:25 AM  #7 
Member Joined: Oct 2013 Posts: 36 Thanks: 0 
bump?hello?

December 30th, 2014, 01:33 PM  #8 
Senior Member Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics 
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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