- #1
karush
Gold Member
MHB
- 3,269
- 5
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...
$y'=\dfrac{1+3x^2}{3y^2-6y},
\quad y(0)=1$
Solving analytically
$3y^2-6y\ dy = 1+3x^2 \ dx$
so far hopefully...