Solutions
- Note that in exercise 1c, the constant function $y=0$ is also a solution to $y^\prime=\frac{1}{x}y+xy^2$.
- In exercise 1b, we have $(z+1)^3=(z^2-1)^3=(z+1)^3(z-1)^3$. To find all solutions $z$, it is important to not simply cancel the two factors of $(z+1)^3$, but instead to rewrite the equation as $0=(z+1)^3\big((z-1)^3-1\big)$. Thus, such a solution satisfies either $z+1=0$ or $(z-1)^3-1=0$.
- For a shortcut in exercise 2, note that
$$\frac{\cos^2(\frac{\pi}{2}x)}{1-x^2}=\frac{1}{2}\frac{\cos(\pi x)+1}{1-x^2}=\operatorname{Re}\left(\frac{1}{2}\frac{e^{i \pi x}+1}{1-x^2}\right).$$
Then, the answer is the real part of
$$\int_{-\infty}^{\infty}\frac{1}{2}\frac{e^{i\pi z}+1}{1-z^2}dz,$$
which can be integrated on a semicircle contour above the real axis.
- Here is a detailed solution for problem 2: Link. The most important part of this problem is the part marked as (iii) in the link.
- The convolution of two functions $f$ and $g$ is a function denoted $f*g$ defined as
$$(f*g)(x)=\int_0^xf(x-x')g(x')dx'.$$
So, in problem 2, we are given
$$x=\int_0^x \frac{1}{\sqrt{x-x'}}y(x')dx'=\frac{1}{\sqrt{x}}*y.$$
We would like to take the Laplace transform of both sides. For this, compute for an exponent $\alpha>-1$ and $s>0$
$$\mathcal{L}(x^\alpha)=\int_0^\infty x^\alpha e^{-sx}dx=\int_0^\infty \left(\frac{u}{s}\right)^\alpha e^{-u} \frac{1}{s}du=\frac{1}{s^{\alpha+1}}\int_0^\infty u^\alpha e^{-u}du=\frac{\Gamma(\alpha+1)}{s^{\alpha+1}}.$$
Thus, taking the Laplace transform of the given equation,
$$\frac{\Gamma(2)}{s^2}=\frac{\Gamma(1/2)}{s^{1/2}}\cdot\mathcal{L}(y).$$
Using $\Gamma(2)=1$ and $\Gamma(1/2)=\sqrt{\pi}$, we have
$$\mathcal{L}(y)=\frac{1}{\sqrt{\pi}s^{3/2}}=\frac{1}{\sqrt{\pi}~\Gamma(3/2)}\mathcal{L}(x^{1/2})=\frac{2}{\pi}\mathcal{L}(\sqrt{x})$$
(we also used $\Gamma(3/2)=\sqrt{\pi}/2$). Finally, taking inverse Laplace transforms gives
$$y(x)=\frac{2}{\pi}\sqrt{x}.$$