# [SOLVED]b.2.2.21 IVP

#### karush

##### Well-known member
Solve the initial value problem
$y'=\dfrac{1+3x^2}{3y^2-6y}, \quad y(0)=1$

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

so far hopefully...

#### Country Boy

##### Well-known member
MHB Math Helper
Well, I would write it as $(3y^2- 6y)dy= (1+ 3x^2)dx$, but, yes, the differential equation can be "separated" so that y is on one side and x on the other.

Now, integrate both sides. Have you done that?

#### karush

##### Well-known member
$y^3 - 3 y^2 =x^3 + x + c$

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

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#### Country Boy

##### Well-known member
MHB Math Helper
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.

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mahalo