# Toronto Math Forum

## MAT244-2013S => MAT244 Math--Tests => Quiz 1 => Topic started by: Victor Ivrii on January 16, 2013, 07:34:46 PM

Title: Night section, 2.2 # 26
Post by: Victor Ivrii on January 16, 2013, 07:34:46 PM

Solve the initial value problem
$$yâ€²=2(1+x)(1+y^2),\qquad y(0)=0$$
and determine where the solution attains its minimal value. -- Added later.
Title: Re: Night section, 2.2 # 26
Post by: Alexander Jankowski 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

$$y(x) = \tan{(x^2 + 2x)}.$$
Title: Re: Night section, 2.2 # 26
Post by: Victor Ivrii 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

\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}

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 $x=-1$â€”provided this point belongs to the domain where solution is defined (all other solutions just shoot from $-\infty$ to $+\infty$ or other way aroundâ€”depending on sign of $x+1$.

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