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
We are given (I think there is an issue in the hint):
$$\tag 1 y = 2y' x + y^3, y^3 \equiv (y')^3$$
If we let $p = \dfrac{dy}{dx}$, we can rewrite $(1)$ as:
$$\tag 2 y = 2px + p^3$$
Differentiating $(2)$ wrt $x$ yields:
$$\dfrac{dy}{dx} = p = 2p + 2 x \dfrac{dp}{dx} + 3 p^2 \dfrac{dp}{dx}$$
This reduces to:
$$\tag 3 3p^2\frac{dp}{dx} + 2x\frac{dp}{dx} + p = 0$$
Solving $(3)$ yields four solutions as:
$$p(x) = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$
From our initial substitution, we have $p = \dfrac{dy}{dx}$, so we can write:
$$\dfrac{dy}{dx} = \pm ~\sqrt{\frac{2}{3}} \sqrt{\pm ~\sqrt{c_1+x^2}-x}$$
We can solve these four (it is possible that not all four satisfy the ODE) for $y(x)$ by rearranging and integrating. For example:
$$\int dy = -\sqrt{\frac{2}{3}} \int \sqrt{~-\sqrt{c_1+x^2}-x}~~dx $$
where $c_1$ is a constant.
Best Answer
$$xy+y+y’^2=0$$ Let : $y’=p$ $$y=-xy’-y’^2=-xp-p^2$$ $$ \frac{dy}{dp} =-p\frac{dx}{dp}-x-2p$$ $$p=y’=\frac{dy}{dx}=\frac{dy}{dp} \frac{dp}{dx} =-p-(x+2p) \frac{dp}{dx}$$ $$2p=-(x+2p) \frac{dp}{dx}$$ $$2p\frac{dx}{dp}+x=-2p$$ The solution of this first order linear ODE is : $$x=-\frac{2}{3}p+\frac{C}{2\sqrt{p}}$$ $$y=-xp -p^2= \frac{2}{3}p^2-\frac{C}{2}\sqrt{p} –p^2 =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p}$$ Finally, the solution of $xy+y+y’^2=0$ expressed on parametric form with parameter $p$ is : $$\begin{cases} x=-\frac{2}{3}p+ \frac{C}{2}\frac{1}{\sqrt{p} } \\ y =-\frac{1}{3}p^2-\frac{C}{2}\sqrt{p} \\ \end{cases}$$ If one want to find the explicit function $y(x)$ the parameter $p$ has to be eliminated from the system $\left(x(p),y(p)\right)$. This is possible, but will lead to complicated equations.