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[SOLVED] 48.2.2.21 IVP

karush

Well-known member
Jan 31, 2012
2,684
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
Jan 30, 2018
426
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
Jan 31, 2012
2,684
$y^3 - 3 y^2 =x^3 + x + c$

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
 
Last edited:

Country Boy

Well-known member
MHB Math Helper
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:

karush

Well-known member
Jan 31, 2012
2,684
mahalo
very helpful

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

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
Notice that the specific condition, |x|< 1, depends also on the fact that the initial condition is at x=0 which is between -1 and 1.

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".