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#### karush

##### Well-known member

- Jan 31, 2012

- 2,684

$y'=\dfrac{1+3x^2}{3y^2-6y},

\quad y(0)=1$

Solving analytically

$3y^2-6y\ dy = 1+3x^2 \ dx$

so far hopefully...

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- Thread starter
- #1

- Jan 31, 2012

- 2,684

$y'=\dfrac{1+3x^2}{3y^2-6y},

\quad y(0)=1$

Solving analytically

$3y^2-6y\ dy = 1+3x^2 \ dx$

so far hopefully...

- Jan 30, 2018

- 426

Now, integrate both sides. Have you done that?

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- #3

- Jan 31, 2012

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$y^3 - 3 y^2 =x^3 + x + c$

ok i think we can plug in $x=0$ and $y=1$ so

$1-3=0+0+c$

$c=-2$

the book ans was

$y^3−3y^2−x−x^3+2=0,\quad |x|<1$

but i don't kmow how they got the interval

ok i think we can plug in $x=0$ and $y=1$ so

$1-3=0+0+c$

$c=-2$

the book ans was

$y^3−3y^2−x−x^3+2=0,\quad |x|<1$

but i don't kmow how they got the interval

Last edited:

- Jan 30, 2018

- 426

That requires some careful thought! The original problem was $y'= \frac{1+ 3x^2}{3y^2- 6y}$. The denominator is $3y^2- 6y= 3y(y- 2)$ so y cannot be 0 or 2. You have, correctly, that $y^3- 3y^2- x^3- x+ 2= 0$. If y= 0, that is $-x^3- x+ 2$ which is 0 when x= 1. If y= 2, that is $8- 12- x^4- x+ 2= -x^3- x- 2$ which is 0 when x= -1. Since y cannot be 0 or 2, x cannot be -1 or 1.

Last edited:

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- Jan 31, 2012

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mahalo

very helpful

when I graphed it i noticed vertical slope at 1 but,,,

very helpful

when I graphed it i noticed vertical slope at 1 but,,,

- Jan 30, 2018

- 426

If we had exactly the same differential equation with condition that y(2) be a given value, since 2> 1, we would have to restrict the solution to "x> 1".

If we had exactly the same differential equation with condition that y(-3) be a given value, since -3< -1, we would have to restrict the solution is "x< -1".