- #1

karush

Gold Member

MHB

- 3,269

- 5

Solve the initial value problem $y'=2y^2+xy^2,\quad y(0)=1$

find min value

$\begin{array}{ll}

\textit{separate variables}

&\dfrac{1}{y^{2}}\ dy=(2+x)\ dx\\ \\

\textit{integrate}

& -\dfrac{1}{y}=2x+\dfrac{x^2}{2}+C\\ \\

\textit{plug in $(0,1)$}

& -\dfrac{1}{1}=0+0+c \therefore c=-1\\ \\

\textit{the equations is}

& y=-1/(x/2+2x-1)+c\\ \\

y'=0 & \textit{returns } x=-2

\end{array}$

typos maybe