MHB -b.2.2.26 IVP min value y'=2(1+x)(1+y^2); y(0)=0

[karush]

368
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 ?

 

[HOI]

By the time you are studying differential equations, you should know that the standard method for finding a minimum, or maximum, value for a function is to set the derivative equal to 0.

Here, you are told immediately that the derivative is $y'= 2(1+ x)(1+ y^2)$. Setting that equal to 0, $2(1+ x)(1+ y^2)= 0$. Since $1+ y^2$ is never 0 we must have x= -1.

 

[HOI]

Add: To check whether x= -1 gives a max or min, use the "second derivative test". $y'= 2(x+ 1)(1+ y^2)$ so $y''= 2(1+ y^2)+ 2(1+ x)(2yy')$. At x= -1 that is $y''= 2(1+ y(-1)^2)$ which is positive no matter what y(-1) is. Therefore x= -1 gives a minimum.

 

[karush]

ok ill try that on the next one