We can transform a second order differential equation of the form $y''=f(t,y')$, using the substitution $v=y', v'=y''$, to a first order equation of the form $v'=f(t,v)$.
Lets do $(3)$ as an example, so we have:
$t^2y''=(y')^2$ , $t>0$
Let: $v=y', v'=y''$, yielding (notice that we have a form $v' = f(t, v)$):
$$t^2 v' = (v)^2$$
We can easily separate this as:
$$\tag 1 \frac{dv}{v^2} = \frac{dt}{t^2}$$
Update
From $(1)$, we get:
$$\frac{1}{v} = \frac{1}{t} + c_1$$
This can be written as: $\displaystyle \frac{1}{v} = \frac{1 + c_1t}{t}$, and then we solve for $v$, yielding:
$$v = \frac{t}{1+c_1t}$$
However, from our earlier substitution, we have $v = y'$, so:
$$y' = \frac{t}{1+c_1t}$$
Can you solve for $y$ now?
Update 2
Note: when you solve for $y'$, you'll get another constant $c_2$ from that, to go along with the $c_1$. You would need to be given some ICs to figure those out - so they are left in this form.
Regards
$$y^{\prime\prime}+\frac{1}{x}y^{\prime}-\frac{x^2+4}{x^2}y=x^4$$
The solution for the associated homogeneous ODE is $y_h=c_1I_2(x)+c_2K_2(x)$
The solution for the non-homogeneous ODE can be found on the form $y=y_h+p(x)$ where $p(x)$ is a particular solution of the ODE.
The seach of a particular solution using the variation parameters method is possible but arduous. Before going on this boring way, it is of use to try some simple functions and see if, by luck, one of them is convenient.
The simplest idea is to try a polynomial. Since there is $-y$ on the left side of the ODE and $x^4$ on the right side, we will try a 4th degree polynomial. Since there is $\frac{-4}{x^2}$ on the left side, the polynomial must not include terms which degree is lower than 2. So, let :
$$p(x)=ax^4+bx^3+cx^2$$
Binging it back into the ODE leads to :
$$p^{\prime\prime}+\frac{1}{x}p^{\prime}-\frac{x^2+4}{x^2}p= -ax^4-bx^3+(12a-c)x^2+5bx=x^4$$
Hence : $a=-1\space;\space b=0\space;\space c=-12$
We see that, "by luck", the polynomial $p(x)=-x^4-12x^2$ is a convenient particular solution. So, the general solution is :
$$y=c_1I_2(x)+c_2K_2(x)-x^4-12x^2$$
Best Answer
Consider the general parametrization $y(x)=x^au(bx^c)$ and insert, \begin{align} y'(x)&=x^{a-1}(au(bx^c)+bcx^cu'(bx^c))\\[.5em] -x^{a-3/2}u(bx^c)=y''(x)&=(a-1)x^{a-2}\Bigl(au(bx^c)+bcx^cu'(bx^c)\Bigr) \\&~~~+x^{a-1}\Bigl((a+c)bcx^{c-1}u'(bx^c)+b^2c^2x^{2c-1}u''(bx^c)\Bigr) \\[.5em] -x^{1/2}u(bx^c)&=a(a-1)u(bx^c)+(2a+c-1)bcx^cu'(bx^c)+b^2c^2x^{2c}u''(bx^c) \end{align}
To get the form of the Bessel equation one needs $c=\frac14$ to satisfy equality of degrees in $x^{1/2}\sim x^{2c}$. Thus inserting with $t=bx^{1/4}$ $$ t^2u''(t)+4(2a-3/4)tu'(t)+16[b^{-2}t^2+a(a-1)]u(t)=0 $$ leading to $b=4$, $8a-3=1\implies a=1/2$, so that $$y(x)=x^{1/2}u(4x^{1/4})~~\text{ and }~~ t^2u(t)+tu(t)+[t^2-4]u(t)=0. $$ As was given in the claim.