Question

hi i have this problem
dy/dx= (x-y)^2
i need the steps ... Thank you .

Answer 1

What is the question?

Answer 2

i need to get y ,,,,,

Answer 3

Hmm I think a substitution will work nicely here,\[\large\rm u=x-y\]

Answer 4

\[\large\rm u'=1-y'\qquad\to\qquad y'=1-u'\]

Answer 5

after this step i will get this function ( du/dx)/(-u^2+1)=1

Answer 6

Me personally I would probably just take ln of both sides

Answer 7

\[\large\rm \frac{1}{1-u^2}du=dx\]Ok looks good so far :)

Answer 8

if the sign +ve then everything great ,, cause it will be tan u ,,, but its -ve :(

Answer 9

Have you learned your Trigonometric Substitutions yet?
If not, you can just look this one up in a table somewhere.

Answer 10

I think it's arcsine or something similar.
\(\large\rm u=\sin\theta\) is the substitution you would make.

Answer 11

Nevermind it's not arcsine like I was thinking XD hehe
ya make the substitution :D

Answer 12

partial fractions will do if the table is not available :)
\[1-u^2 = (1-u)(1+u)\]

Answer 13

\[\large\rm \int\limits\frac{1}{1-u^2}du=\int\limits dx\]
\[\large\rm u=\sin \theta,\qquad\qquad\qquad du=\cos \theta~d \theta\]
\[\large\rm \int\limits\limits\frac{1}{1-\sin^2\theta}\cos \theta ~d \theta=\int\limits\limits dx\]Yes, partial fractions is another good option c:

Answer 14

this is homework to master students ,,, I think its not as easy as it seems .

Answer 15

i know the answer but i can't link between the answer and the question in steps .

Answer 16

Hmm, ya it's quite a few steps after the trig sub even...
hmm this one is a bit of a doozy XD

Answer 17

http://www.wolframalpha.com/input/?i=dy%2Fdx%3D(x-y)%5E2

Answer 18

this should help in imagin the answer ,,, but it still not easy . :(

Answer 19

I'll just show you steps :D
If they don't make sense uhhh hmm

Answer 20

\[\large\rm \int\limits\limits\limits\frac{1}{\cos^2\theta}\cos \theta ~d \theta=\int\limits\limits\limits dx\]Cancelling out a cosine,\[\large\rm \int\limits\limits\limits\frac{1}{\cos \theta}~d \theta=\int\limits\limits\limits dx\]rewrite as secant,\[\large\rm \int\limits\limits\limits\sec\theta~d \theta=\int\limits\limits\limits dx\]Integrate,\[\large\rm \ln|\sec \theta+\tan \theta|=x+c\]

Answer 21

Exponentiate each side,\[\large\rm e^{\ln|\sec \theta+\tan \theta|}=e^{x+c}\]simplify,\[\large\rm \sec \theta+\tan \theta=Ce^x\]

Answer 22

Use our trig sub to set up a triangle which will help us undo our substitution.\[\large\rm \sin \theta=u\qquad\to\qquad \sin \theta=\frac{opposite}{hypotenuse}=\frac{u}{1}\]

Answer 23

Pythagorean Theorem to find the missing side,

Answer 24

Now we can undo our trig sub,\[\large\rm \sec \theta+\tan \theta=Ce^x\]with what we know about the side relations,\[\large\rm \frac{hypotenuse}{adjacent}+\frac{opposite}{adjacent}=Ce^x\]and our triangle,\[\large\rm \frac{1}{1-u^2}+\frac{u}{1-u^2}=Ce^x\]

Answer 25

\[\large\rm \frac{1+u}{1-(u)^2}=Ce^x\]Undoing our u-substitution,\[\large\rm \frac{1+x-y}{1-(x-y)^2}=Ce^x\]

Answer 26

And thennnnn some more work from there to isolate our y +_+
ughhhhh

Answer 27

Multiply through by the denominator,\[\large\rm 1+x-y=Ce^x-Ce^x(x-y)^2\]Expanding the square... completing the square.. etc etc :O
Do I have to keep going grrr

Answer 28

llol

Answer 29

this great idea ,, with great thinking ,,, Thank you ... i will complete the steps ,, and if i have troubles i will send you message :)

Answer 30

cool c:

Answer 31

the result i got from this reduction is \[y=x-\frac{ ce^x+1 }{ce^x }\]

Answer 32

Hmmm ya, I"m wondering if I made a boo boo somewhere :o
I can't seem to solve for y in an easy way.
I don't think your form turned out quite right, hmm

Answer 33

your steps very clever ,,, i believe its correct .

Answer 34

\[y=x-\frac{ e^{x+c}+1 }{e^{x+c} }\]

Answer 35

the solution can be in this form ,, you made reduction by consider it = to C ,,, i made it e^c ,,, because it give sense to the answer .

Answer 36

\[\int\limits_{}^{}\frac{ \frac{ du }{dx } }{-u^2+1 }dx=-\frac{ 1 }{ 2 }\log(-u+1)+\frac{ 1 }{ 2 }\log(u+1)\]

Answer 37

How ???!!!! @zepdrix

Answer 38

\[\large\rm 1-u^2=(1-u)(1+u)\]Have you learned about Partial Fraction Decomposition?\[\large\rm \frac{1}{(1-u)(1+u)}\quad=\frac{A}{1-u}+\frac{B}{1+u}\]

Answer 39

its \[\frac{ 1+U }{(1-U^2) }=\frac{ 1 }{1-U }\]