- #1

karush

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MHB

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Solve the IVP $y'=2(1+x)(1+y^2),\quad y(0)=0$

$\begin{array}{ll}

\textit{separate variables}&

\displaystyle

\left(\dfrac{1}{1+y^2}\right)\ dy

=2(1+x)\ dx\\

\textit{integrate thru}&

\arctan \left(y\right)=2x+x^2+c\\

\textit{plug in x=0 and y=0}&

\arctan 0=0+c\\

&0=c\\

\textit{thus the equation is}&

y=\tan(2x-x^2)

\end{array}$

find where solution attains minimum value

ok wasn't sure about the tangent thing

book answer for min value is -1 but ?

Solve the IVP $y'=2(1+x)(1+y^2),\quad y(0)=0$

$\begin{array}{ll}

\textit{separate variables}&

\displaystyle

\left(\dfrac{1}{1+y^2}\right)\ dy

=2(1+x)\ dx\\

\textit{integrate thru}&

\arctan \left(y\right)=2x+x^2+c\\

\textit{plug in x=0 and y=0}&

\arctan 0=0+c\\

&0=c\\

\textit{thus the equation is}&

y=\tan(2x-x^2)

\end{array}$

find where solution attains minimum value

ok wasn't sure about the tangent thing

book answer for min value is -1 but ?

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