Question

Help me out with this ....

Answer 1

The solution of $$\frac{dy}{dx}=\frac{(x+y)^2}{(x+2)(y-2)}$$. I attempted by substituting$$\ y-2=t$$ but not able to find the solution. The options are

$$\ a.(x+2)^4(1+\frac{2y}{x})=ke^{\frac{2y}{x}}$$

$$\ b. (x+2)^4(1+\frac{2(y-2)}{(x+2)})=ke^{\frac{2(y-2)}{x+2)}}$$

$$\ c.(x+2)^3(1+\frac{2(y-2)}{(x+2)})=ke^{\frac{2(y-2)}{x+2)}}$$

d. None of the above.

Answer 2

@ganeshie8

Answer 3

http://math.stackexchange.com/questions/152842/fracdydx-1-frac2xy-solution-by-an-integrating-term?rq=1

thought if it is like the ^^^^^^^^^^^^^^^, but is much out of phase than what i need.

Answer 4

thats a very good try! and you're almost there

Answer 5

try this :

y-2 = t
x+2 = u

Answer 6

dy/dx = dt/du

Answer 7

the eq'n becomes :
\[\frac{dt}{du}=\frac{(u+t)^2}{ut}\]

Answer 8

should i expand the numerator ????

Answer 9

nope, its a homogeneous equation

familiar with solving homogeneous equations ?

Answer 10

Ohh I didnt, notice it, ok so u/t = v and will solve the rest.. thank you :)

Answer 11

t/u=v

Answer 12

try t = uv

Answer 13

yes!
t is the dependent variable, so we need to mess with it

Answer 14

$$u=v-log(v+1)+logc$$, where log c is constant.

Answer 15

im getting a different solution

Answer 16

\(\rm t = uv \implies \frac{dt}{du}= u\frac{dv}{du} + v\)

Answer 17

ya I got that .. !!

Answer 18

\(\large \rm\frac{dt}{du}=\frac{(u+t)^2}{ut}\)

becomes :

\(\large \rm u\frac{dv}{du}+v=\frac{(1+v)^2}{v}\)

Answer 19

sednding that v to right hand side and sinplifying gives u :

\(\large \rm u\frac{dv}{du}=\frac{1+2v}{v}\)

Answer 20

separate variables now

Answer 21

stupid me!!! I wrote v instead of 2v in the numerator expansion.

Answer 22

i see... lets finish it off

Answer 23

\(\large \rm u\frac{dv}{du}=\frac{1+2v}{v}\)


\(\large \int \rm \frac{v}{2v+1}dv=\int \frac{1}{u}du\)

Answer 24

\(\large \int \rm \frac{1}{2}\frac{2v+1-1}{2v+1}dv=\ln |u| + C\)

Answer 25

\(\large \rm \frac{1}{2} \int 1-\frac{1}{2v+1}dv=\ln |u| + C\)

Answer 26

\(\large \rm \frac{1}{2} v - \frac{1}{4}\ln | 2v+1|=\ln |u| + C\)

Answer 27

does that look correct ?

Answer 28

I think it will be 1/2 ln|2v+1| cause its d/dx is 1/2*(2/2v+1) ?????

Answer 29

sorry d/dv

Answer 30

there is a 1/2 factor outside the integral already

Answer 31

oh!! yes. so the solution is $$\large \rm \frac{1}{2} v - \frac{1}{4}\ln | 2v+1|=\ln |u| + C$$, I will put the values in terms of x and y and extract the solution.

Thank you :)

Answer 32

good, maybe combine that C into ln and write it as ln|Cu|

Answer 33

okay !!!

Answer 34

:) Thanks and Good night :)

Answer 35

Night :)

Answer 36

$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$

Help me with this @ganeshie8

Answer 37

you're done with the other problem ?

Answer 38

I have used the reduced homogenous equation process and reached the following

$$2tan^{-1}x=\int{\frac{a^2-u}{u}}du$$

but not able to do the LHS integration help me

Answer 39

yes I did the other one, $$f(x)=\sqrt{tan2x}$$ and when I used the limits I found the option b is correct i.e $$b-a\le \frac {\pi}{4}$$