Author Topic: Night section, 2.2 # 26  (Read 3758 times)

Victor Ivrii

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Night section, 2.2 # 26
« on: January 16, 2013, 07:34:46 PM »
Please post solution

Solve the initial value problem
$$y′=2(1+x)(1+y^2),\qquad y(0)=0$$
and determine where the solution attains its minimal value. -- Added later.
« Last Edit: January 18, 2013, 03:29:18 PM by Victor Ivrii »

Alexander Jankowski

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Re: Night section, 2.2 # 26
« Reply #1 on: January 16, 2013, 10:29:13 PM »
The given first-order nonlinear ordinary differential equation is separable, so

$$
\frac{dy}{dx} = 2(1+x)(1+y^2) \Rightarrow \frac{dy}{1+y^2} = 2(1+x)dx \Rightarrow \arctan{y} = x^2 + 2x + C \Leftrightarrow y(x) = \tan{(x^2 + 2x + C)}.
$$

Using the initial condition, we find $C$:

$$
0 = \tan{(0^2 + 2(0) + C)} \Leftrightarrow C = \arctan{0} = 0
$$

Conclusively, the solution to the initial value problem is

$$
y(x) = \tan{(x^2 + 2x)}.
$$

Victor Ivrii

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Re: Night section, 2.2 # 26
« Reply #2 on: January 18, 2013, 03:49:06 PM »
In the perfect solution one should determine where solution is defined. Obviously it happens where $-\frac{\pi}{2}<2x +x^2<\frac{\pi}{2}$ (as other intervals do not contain $x=0$) and resolving one arrives to interval
\begin{equation}
\left(-\frac{1}{2}-\sqrt{\frac{\pi}{2}+\frac{1}{4}},  -\frac{1}{2}+\sqrt{\frac{\pi}{2}+\frac{1}{4}}\right).
\label{eq-1}
\end{equation}

Also I forgot the last part in online version: "and determine where the solution attains its minimal value".

While one can find minimum of the found solution directly one can also observe from equation that $y'=0$ as $x=-1$ and $y'>0$ as $x>-1$, and $y'<0$ as $x<-1$ so solution attains its minimal value as $x=-1$—provided this point belongs to the domain where solution is defined (all other solutions just shoot from $-\infty$ to $+\infty$ or other way around—depending on sign of $x+1$.


Here I am discussing the perfect solution not the grading criteria.