Question

dy/dx=sin(x+y)

Answer 1

this is a bit** of a DE

Answer 2

Lol...wat u have to find here

Answer 3

y(x)

Answer 4

x+y=k...implies..dy/dx+1=dk/dx....dk/dx=sink-1....dk/(sin k-1)=dx....solve by using tn k/2

Answer 5

i got

∫1/(sin(k)+1)dk=∫dx
=x+c


i dont know how to integrate

∫1/(sin(k)+1) dk

Answer 6

ok...so 2 tan k/(1+tan^2)= sin k...now try it...and see if it works

Answer 7

sorry ..tan k/2

Answer 8

oh...nd tan^k/2..typing mistke

Answer 9

did u get it uncle??

Answer 10

@UnkleRhaukus do I need to elaborate..??

Answer 11

weierstrass substitution for 1/ (1+sin x)
t= tan (x/2)

Answer 12

Unkle, you have solved way more difficult integrals than this :)

Answer 13

∫ 2 / (t+1)^2 dt

Answer 14

:D..so did u...

Answer 15

=-2 / (tan (k/2)+1) ?

Answer 16

sry..so u did it..

Answer 17

= -2 / [tan((x+y)/2)+1] ?

Answer 18

i haven't worked it out...

Answer 19

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

Answer 20

is this right?

Answer 21

Yup..

Answer 22

unkle do u like number theory?

Answer 23

yes, how did you check my solution?

Answer 24

maybe i should have kept the 2 in the numerator,

Answer 25

\[\newcommand \p \newcommand
\p \dd [1] { \,\mathrm d#1 } % infinitesimal
\p \de [2] { \frac{\mathrm d #1}{\mathrm d#2} } % first order derivative
\begin{align} \vdots \\
\int\frac{2\dd t}{(t+1)^2} &=x+c \\
& &\text{let u}&=t+1 \\
& & \dd u &=\dd t \\
\int \frac2{u^2}\dd u &=x+c \\
\frac2{-u} &=x+c \\
-\frac2{t+1} &=x+c \\
-\frac2{\tan\big(\frac w2\big)+1} &=x+c \\
-\frac2{\tan\big(\frac{x+y}2\big)+1} &=x+c
\end{align}\]