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[SOLVED] 48.2.2.23 IVP and min valuew

karush

Well-known member
Jan 31, 2012
2,678
\tiny{b.48.2.2.23}
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:cool:
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
There is a typo where you have
$y= -1/(x/2+ 2x- 1)+ c$
You clearly intended "$x^2/2$". Also, having determined that the constant of integration was -1, where did that additional "c" come from?
 

karush

Well-known member
Jan 31, 2012
2,678
here is the book answer

2020_07_18_14.14.32_edit.jpg
yeah your right about the $x^2$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
426
\tiny{b.48.2.2.23}
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:cool:
Every thing is good until that last equation!
Integrating gives
$-\frac{1}{y}= \frac{x^2}{2}+ 2x+ C$
an y(0)= 1 gives C= -1 so
$-\frac{1}{y}= \frac{x^2}{2}+ 2x- 1$.

Then solving for y (if you are asked to do so- I would consider the previous line a perfectly good solution)
$y= \frac{-1}{\frac{x^2}{2}+ 2x- 1}$
or, if you don't like a fraction in the denominator of afraction, multiply both numerator and denominator by 2:
$y= \frac{-2}{x^2+ 4x- 2}$

But, again, where did that "+c" in your last line come from?