Question

\[\frac{\mathrm dy}{\mathrm dx}=\sin(x+y)\]

Answer 1

\[\newcommand\p \newcommand
\p \dd [1] { \,\mathrm d#1 }
\p \de [2] { \frac{ \mathrm d #1}{\mathrm d#2} }
\begin{align} \de yx &=\sin(x+y) & y(0) &=0 \\
\\
& &\text{let } w&=x+y \\ & & \de wx&=1+\de yx \\
\de wx-1 &=\sin w \\
\de wx &=\sin w+1 \\
\frac{\dd w}{\sin w+1} &=\dd x \\
\int\frac{\dd w}{\sin w+1} &=\int\dd x & &\text{Weierstrass substitution} \\
& &\text{let }t&=\tan\big(\frac w2\big) \\
& &\frac{2\dd t}{1+t^2}&=\dd w \\
& & \frac{2t}{1+t^2}&=\sin w \\
\int\frac1{\frac{2t}{1+t^2}+1}\frac{2\dd t}{1+t^2} &=\int\dd x \\
\int\frac{2\dd t}{2t+1+t^2} &=x+c \\
\int\frac{2\dd t}{(t+1)^2} &=x+c \\
& &\text{let u}&=t+1 \\
& & \dd u &=\dd t \\
\int \frac2{u^2} \dd u &=x+c \\
\frac2{-u} &=x+c \\
-\frac2{t+1} &=x+c \\
-\frac2{\tan\big(\frac w2\big)+1} &=x+c \\
-\frac2{\tan\big(\frac{x+y}2\big)+1} &=x+c & c&=-2 \\
\frac1{\tan\big(\frac{x+y}2\big)+1} &=1-x/2 \\
\tan\Big(\frac{x+y}2\Big)+1 &=\frac1{1-x/2} \\
\tan\Big(\frac{x+y}2\Big) &=\frac{1-(1-x/2)}{1-x/2} \\
\frac{x+y}2 &=\arctan\Big(\frac x{x-2}\Big) \\
y(x) &=2\arctan\Big(\frac x{x-2}\Big)-x
\end{align}
\]

Answer 2

Well done! *claps* :)

Answer 3

i think something is wrong

Answer 4

im compairing to a forward euler scheme for approximation , and it looks very different

Answer 5

the technique is fine ... for faster just use this
\[ \int \frac{1}{(\sin (x/2) + \cos(x/2))^2}dx = 2\frac{\sin(x/2)}{\sin(x/2) + \cos(x/2)} \]

1/(sin(x) + cos(x))^2 = (sin^2(x) + cos^2(x))/1/(sin(x) + cos(x))^2 = cos(x)(sin(x) + cos(x)) - sin(x)(cos(x) - sin(x))/1/(sin(x) + cos(x))^2
Euler approximation is just an approximation. there is always an error associated with it unless your solution is linear. Use Runge-Kutta methods if you want to get more precise solution.

Answer 6

1 + sin(x) = (sin(x/2) + cos(x/2))^2

Answer 7

heh that is weird
http://www.wolframalpha.com/input/?i=solve+y%27+%3D+sin%28x%2By%29+using+euler%27s+method+with+y%280%29%3D0%2C+h%3D0.15

Answer 8

@mukushla

Answer 9

numerical solution contradicts exact solution completely !!!

Answer 10

where am i going wrong ?

Answer 11

for some reason i can't even get my matlab plot for the exact value to hit the right intercept

Answer 12

it's weird, did u see the wolf answer??

Answer 13

even forth order runge-kutta !!!!

Answer 14

changin the type of method wont help here, to improve accuracy is eaisier to change the step size, but thats not the problem i am having.

my appproximation curve using the euler method is fine,

it my exact analtic solution that is wrong

Answer 15

let me check the exact solution. i'll let you know if i get something different.

Answer 16

i'm getting\[y=2 \tan^{-1} \frac{x}{2-x} -x\]

Answer 17

that is good

Answer 18

in your solution consider last three line, going from first to second it will be \(\frac{x}{2-x}\) not \(\frac{x}{x-2}\)

Answer 19

rest is fine, i'm curious about the plots now, would you please show the plots

Answer 20

yeah i still am getting something strange in matlab

Answer 21

a differnt graphing program is getting the right results now though

Answer 22

i can't figure out what's wrong with matlab program :D ???!!! while other one is right !!

Answer 23

i just found that
```
>> ezplot('y=2*atan(x/(2-x))-x',[0 1])
```
works

Answer 24

im not experienced with matlab, but i seem that `plot` and `ezplot` are different computation methods

Answer 25

i kinda wanted to be able to just `plot`,

Answer 26

i bet the problem with `plot` in matlab is the same as the problem that wolfram is having

Answer 27

i lost connection, yeah they must be same !!

Answer 28

well ive got this now, its working alright,
i'd still like some kinda explaination as to what went wrong