Question

Diffrential Eq. Q..

Answer 1

\[\frac{ dy }{ dx }=\frac{ y^3 }{ 2(xy^2-x^2 )}\]

Answer 2

find the solution..

Answer 3

@zepdrix @imqwerty @rvc

Answer 4

Try to take all the y variables on the l.h.s and all the x variables alone on l.h.s

Answer 5

u mean x on rhs?

Answer 6

yes all the x variables on rhs

Answer 7

it doesnt matter, just y on different side and x on different side

Answer 8

i am not able reduce after this..
\[\frac{ y^2 - x }{ y^3 } dy=\frac{ dx }{ 2x } \]

Answer 9

?

Answer 10

\[y^2e(\frac{ -y^2 }{ x })\]

Answer 11

this is what is given..it has a printing mistake

Answer 12

anyway what was ur answer ?

Answer 13

no..this is the question..i know how to solve if it was x^3..it would be much simpler!

Answer 14

are we given any info? I can try to test for if it can be written as an exact equation

Answer 15

z=y^2
dz/dx=2y dy/dx
\(\Huge \frac{ dy }{ dx }=\frac{ y^3 }{ 2(xy^2-x^2 )}\)
\(\Huge \frac{ dy }{ dx }=\frac{ zy}{ 2(xz-x^2 )}\)
\(\Huge \frac{ dz }{ dx }/(2y)=\frac{ zy}{ 2(xz-x^2 )}\)
\(\Huge \frac{ dz }{ dx }=\frac{ z^2}{ (xz-x^2 )}\)
your equation is NOW homogenous. Divide both numerator and denominator by x^2; you know where to take it from here.

Answer 16

\[\frac{ dz }{ dx }= \frac{ (z/x)^2 }{ 2(z/x - 1)}\]

Answer 17

i m not sure how to proceed.. @inkyvoyd

Answer 18

take v=z/x, then xv=z and z'(x)=v'x+v, and then your equation becomes
\(\Huge(v'x+v)=\frac{v^2}{v-1}\)

Answer 19

ok got it! thanks !

Answer 20

yeah don't forget to make all those back subsitutions (and if you don't get it yet this equation is SEPARABLE)

Answer 21

i got it!

Answer 22

i am getting
\[logy=y^2/x +c\]

Answer 23

tbh I did not work out the problem lol... does that not match the answer?

Answer 24

no..i have written the answer abv..it has a printing mistake though

Answer 25

well the solution to an equation is an equation, and you've given an expression... what's on the other side? zero?

Answer 26

thats y i said it has a printing mistake..also coz there is no "power" for e..

Answer 27

can u just work out the problem and tell me the answer..?(its just 3 steps)

Answer 28

\(\Huge(v'x+v)=\frac{v^2}{v-1}\)
\(\Huge v'x=\frac{v^2}{v-1}-\frac{v^2-v}{v-1}\)
\(\Huge v'x=\frac{v}{v-1}\)
\(\Huge \frac{v-1}{v}v'=1/x\)
\(\Huge v-\ln|v|=\ln|x|+C\)
\(\Huge z/x-\ln|z/x|=\ln|x|+C\)
\(\Huge y^2/x-\ln|y^2/x|=\ln|x|+C\)

Answer 29

oh yeah made a small mistake..now fine!

Answer 30

\(\Huge y^2/x-\ln|y^2/x|=\ln|x|+C\)
\(\Huge y^2/x=\ln|y^2/x|+\ln|x|+C\)
\(\Huge y^2/x=\ln|y^2|+C\)
\(\Huge y^2=Ce^{y^2/x}\)

Answer 31

and i figured out the printing mistake

Answer 32

oh thats what u have written above!!

Answer 33

no...i think u have made a mistake

Answer 34

lol, I'm pretty sure I made a mistake too, this is why I did not try to solve the equation this late at night ;)

Answer 35

its ok! np!

Answer 36

Thanks once again!

Answer 37

best wishes :)

Answer 38

thanks!

Answer 39

i have one more q..

Answer 40

post it haha

Answer 41

ya..wait..

Answer 42

solving x seems to be easier here..