UnkleRhaukus
  • UnkleRhaukus
dy/dx=sin(x+y)
Differential Equations
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
this is a bit** of a DE
anonymous
  • anonymous
Lol...wat u have to find here
UnkleRhaukus
  • UnkleRhaukus
y(x)

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More answers

amriju
  • amriju
x+y=k...implies..dy/dx+1=dk/dx....dk/dx=sink-1....dk/(sin k-1)=dx....solve by using tn k/2
UnkleRhaukus
  • UnkleRhaukus
i got ∫1/(sin(k)+1)dk=∫dx =x+c i dont know how to integrate ∫1/(sin(k)+1) dk
amriju
  • amriju
ok...so 2 tan k/(1+tan^2)= sin k...now try it...and see if it works
amriju
  • amriju
sorry ..tan k/2
amriju
  • amriju
oh...nd tan^k/2..typing mistke
amriju
  • amriju
did u get it uncle??
amriju
  • amriju
@UnkleRhaukus do I need to elaborate..??
hartnn
  • hartnn
weierstrass substitution for 1/ (1+sin x) t= tan (x/2)
hartnn
  • hartnn
Unkle, you have solved way more difficult integrals than this :)
UnkleRhaukus
  • UnkleRhaukus
∫ 2 / (t+1)^2 dt
amriju
  • amriju
:D..so did u...
UnkleRhaukus
  • UnkleRhaukus
=-2 / (tan (k/2)+1) ?
amriju
  • amriju
sry..so u did it..
UnkleRhaukus
  • UnkleRhaukus
= -2 / [tan((x+y)/2)+1] ?
amriju
  • amriju
i haven't worked it out...
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand \p \newcommand \p \dd [1] { \,\mathrm d#1 } % infinitesimal \p \de [2] { \frac{\mathrm d #1}{\mathrm d#2} } % first order derivative \begin{align} \de yx &=\sin(x+y) & y(0) &=0,\qquad x=[0,1] \\ \\ \\ & &\text{let } w&=x+y \\ & & \de wx&=1+\de yx \\ \de wx-1 &=\sin w \\ \de wx &=\sin w+1 \\ \frac{\dd w}{\sin w+1} &=\dd x \\ \int\frac{\dd w}{\sin w+1} &=\int\dd x &&\text{Weierstrass substitution} \\ & &\text{let }t&=\tan\frac w2 \\ & &\frac{2\dd t}{1+t^2}&=\dd w \\ & & \frac{2t}{1+t^2}&=\sin w \\ \int\frac{\dd w}{\frac{2t}{1+t^2}+1}\frac{2\dd t}{1+t^2} &=\int\dd x \\ \int\frac{2\dd t}{2t+1+t^2} &=x+c \\ \int\frac{2\dd t}{(t+1)^2} &=x+c \\ & &\text{let u}&=t+1 \\ & & \dd u &=\dd t \\ \int \frac2{u^2}\dd u &=x+c \\ \frac1{-u} &=x+c \\ -\frac1{t+1} &=x+c \\ -\frac1{\tan\big(\frac w2\big)+1} &=x+c \\ -\frac1{\tan\big(\frac{x+y}2\big)+1} &=x+c \end{align}\]
UnkleRhaukus
  • UnkleRhaukus
is this right?
amriju
  • amriju
Yup..
amriju
  • amriju
unkle do u like number theory?
UnkleRhaukus
  • UnkleRhaukus
yes, how did you check my solution?
UnkleRhaukus
  • UnkleRhaukus
maybe i should have kept the 2 in the numerator,
UnkleRhaukus
  • UnkleRhaukus
\[\newcommand \p \newcommand \p \dd [1] { \,\mathrm d#1 } % infinitesimal \p \de [2] { \frac{\mathrm d #1}{\mathrm d#2} } % first order derivative \begin{align} \vdots \\ \int\frac{2\dd t}{(t+1)^2} &=x+c \\ & &\text{let u}&=t+1 \\ & & \dd u &=\dd t \\ \int \frac2{u^2}\dd u &=x+c \\ \frac2{-u} &=x+c \\ -\frac2{t+1} &=x+c \\ -\frac2{\tan\big(\frac w2\big)+1} &=x+c \\ -\frac2{\tan\big(\frac{x+y}2\big)+1} &=x+c \end{align}\]

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