Question

solve dy/dx=6x/(y^2(3-x^2)) for y(1)=1

Answer 1

do you know variable separable method ?
did you try to separate the variables for this DE ?

Answer 2

yep, I got \[- \sqrt[3]{3 \ln (3-x ^{2})-3c}\]

Answer 3

is that right?

Answer 4

ok, now you can use the info : y(1) =1

which means , when x= 1, y = 1

you can plug these in the equation you got , to get the value of c

Answer 5

let me check

y^3/3 = -3 ln (3-x^2)+ c

is that what you got first ?

Answer 6

I got c= \[3\ln (2)+1/3\]

Answer 7

you can keep the equation as

y^3/3 = -3 ln (3-x^2)+ c
to get the value of c

Answer 8

yeah, that's what I got

Answer 9

so
1/3 = -3 ln 2 +c
yes, your 'c' is correct!

Answer 10

really? i wasn't sure.. alrighty then! thank you!

Answer 11

welcome ^_^
did you correctly got your final answer ?

Answer 12

well, for the final answer, I got y = \[- (9 \ln (\frac{ 3-x ^{2} }{ 2 }) + 1)^{\frac{ 1 }{ 3 }}\]

Answer 13

yeah, not entirely sure for that bit

Answer 14

let me reshuffle terms to see whether they can be simplified


y^3/3 = -3 ln (3-x^2)+ 3 ln 2 +1/3

(y^3-1) = -9 [ln (3-x^2) -ln 2 ]

y^3 -1 = -9 ln [(3-x^2)/2]

y^3 = 1-9ln [(3-x^2)/2]

\(\Large y = \sqrt[3]{1-9\ln (\dfrac{3-x^2}{2})}\)

Answer 15

looks like you almost got the same thing, just be careful with the brackets

Answer 16

\( (-9 \ln (\frac{ 3-x ^{2} }{ 2 }) + 1)^{\frac{ 1 }{ 3 }}\)

Answer 17

ah yeah, I think that's why I got the second part wrong.. it asks to solve dy/dx - x^2y - xy = 0.. Thanks! :)

Answer 18

welcome ^_^

Answer 19

dy/dx = y (x^2-x)

again variable seperable

1/y dy = (x^2-x) dx
piece of cake from here :P

Answer 20

sorry to bother again but I can't seem to solve the IVP \[dy/dx - x^{2}y - xy = 0\]

Answer 21

add x^2y + xy on both sides first

Answer 22

ahh ok, and solve it from there, yeah? i see it... i'll try it now :)

Answer 23

y is the equation I found before yeah?

Answer 24

i thought this one is completely new and different question....

Answer 25

oh sorry, it's not.. it's a continuation of the previous one

Answer 26

thats confusing...anyways, keep y as y only, as of now...

factor out y from right side

Answer 27

\[\frac{ 6x }{ y(3-x ^{2}) } = x ^{2} + x\] am I on the right track?

Answer 28

hold on.. i think i'm wrong

Answer 29

okay, so dy/dx = y(x^2 + x)

Answer 30

you need to solve to find expression for 'y' again , in terms of x, right ?

Answer 31

yep

Answer 32

ok, then separate the variables now

Answer 33

dy/dx = y(x^2 + x)

(1/y)dy = (x^2+x) dx

and integrate both sides

Answer 34

In(y) = (x^3)/3 + (x^2)/2

Answer 35

correct.

don't forget the +c

Answer 36

oh crap.. yep, my bad. thanks

Answer 37

welcome ^_^

Answer 38

umm.. then how would I continue?

Answer 39

\(\large \ln a = b \implies a = e^b \)

Answer 40

alright..

Answer 41

In(y) = (x^3)/3 + (x^2)/2 = x^2/6 (2x+3) + c

\(\large y = c e^{[\dfrac{x^2}{6} (2x+3)]}\)

Answer 42

\[y = e^{\frac{ x^3 }{ 3 } + \frac{ x^2 }{ 2 } + c}\]

Answer 43

ok, then you can use
\(e^{a+b} = e^a\times e^b\)

so you'll get e^c, whic is also a constant, and you can write it as c only

Answer 44

\(\Large y = e^{\frac{ x^3 }{ 3 } + \frac{ x^2 }{ 2 } + c} \implies y = e^{\frac{ x^3 }{ 3 } + \frac{ x^2 }{ 2 } }\times e^c \\ \Large y =c e^{\frac{ x^3 }{ 3 } + \frac{ x^2 }{ 2 }} \)

Answer 45

if you want, you can simplify the exponent

Answer 46

yep! I got that too :)

Answer 47

if you simplify it, you'll get
\(\large y = c e^{[\dfrac{x^2}{6} (2x+3)]}\)

Answer 48

oh alright then

Answer 49

and then would I use the the y(1) = 1 to solve for c?

Answer 50

if y(1) = 1 is also given for this question, then yes

Answer 51

c=e^(-5/6)

Answer 52

thats correct :)

Answer 53

so \[y = e^\frac{ -5 }{ 6 } \times e^ (\frac{ x^2 }{ 6 } (2x+3))\]

Answer 54

yeah, you can combine the exponents
\(\large y = e^{[\dfrac{x^2}{6} (2x+3)]-\dfrac{5}{6}} =e^{\dfrac{x^2(2x+3)-5}6} \)

Answer 55

ahh yep, got that :)