Toronto Math Forum
MAT2442013S => MAT244 MathTests => Quiz 1 => Topic started by: Victor Ivrii on January 16, 2013, 07:34:46 PM

Please post solution
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.

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

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{eq1}
\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 $x=1$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.