Question

Given that y = 1 when x = 0, solve the differential equation dy dx = y3 + 1 y2 , obtaining an expression for y in terms of x.

Answer 1

sorry it is dy/dx = (y^3 +1)/y^2

Answer 2

\[dy/dx = \frac{(y^{3} + 1)} {y^{2}}\]
is that it?

Answer 3

yes

Answer 4

lol i feel silly but it's a no solution XD look at Dy/Dx X being zero, you can't divide by zero :P

Answer 5

[y^2 / (y^3+1)] dy = dx
integrating both sides
y^3 + 1 = u
du = 3y^2dy
du/3 = y^2dy

du / (3u) = dx
x +C = (1/3) ln|u|
x +C = (1/3) ln|y^3 + 1|

3e^(x+c) -1 = y^3
y = (3e^(x+c) - 1)^(1/3)

y(0) = 1
1 = (3e^c -1)^(1/3)
1 = 3e^c - 1
2/3 = e^c
c = ln(2/3)

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

Answer 6

hope i havent done any mistakes here though

Answer 7

\[\frac{ dy }{ dx } = \frac{ (y^3+1) }{ y^2 } = \frac{ y^3 }{ y^2 } + \frac{ 1 }{ y^2 }\]\[\frac{ dy }{ dx } = y + y^-2\]\[y = \int\limits_{}^{}(y + y^-2) dx\]

Answer 8

\[\frac{ DY }{ DX } <-- cannot divide by zero\]

Answer 9

@Coolsector answer is like [2e^(3x)-1]^(1/3)

Answer 10

let me check again

Answer 11

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

Answer 12

my answer is the same .. we just had to simplify it

Answer 13

because e^(ln(2/3)) = 2/3 :)

Answer 14

@Coolsector oh ok haha thanks :D

Answer 15

yw

Answer 16

@Coolsector

how come this.. du / (3u) = dx
becomes this? x +C = (1/3) ln|u| ?
plz explain thanks :)

Answer 17

@gaara you dont know any calculus do you?

Answer 18

@smoothieoeek
\[\int\limits_{}^{}\frac{du}{3u} = \frac{1}{3}\int\limits_{}^{}\frac{du}{u} = \frac{1}{3} \ln |u|\]
and
\[\int\limits_{}^{}dx = x+C\]