I think that it's intended that they are all solved using the so called Integrating Factor (IF) Method.
The standard form of the equation is taken to be
\(\displaystyle \frac{dy}{dx}+P(x)y = Q(x)\) ,
and the method of solution is to multiply throughout by the IF \(\displaystyle e^{\int P(x)\,dx}\). That will put the LHS into the form of a perfect derivative.
For example, with the second one, \(P(x)=\tan(x) \text{ so the IF will be } \exp(\ln(\sec(x))=\sec(x)\).
That's what CPhill multiplied by. (I don't know whether he worked it out as above or whether it was by inspection).
For the first one, \(\displaystyle x\frac{dy}{dx} +2y=x^{2}-3\),
divide throughout by x to begin with to put the equation into its standard form, so we have
\(\displaystyle \frac{dy}{dx}+\frac{2}{x}y = x -\frac{3}{x}\) and the IF will be \(\displaystyle \exp(\int \frac{2}{x}\;dx)=\exp(2\ln(x))=\exp(\ln(x^{2}))=x^{2}\).
Multiply throughout by \(\displaystyle x^{2}\) and the equation becomes
\(\displaystyle x^{2}\frac{dy}{dx}+2xy=x^{3}-3x\)
which can be written as
\(\displaystyle \frac{d}{dx}(x^{2}y)=x^{3}-3x\),
and now just integrate wrt x.
The third one also begins with a division by x.
The fourth one requires a knowledge of hyperbolic functions. The IF will be \(\displaystyle e^{\;\sinh^{-1}x}\) and we have to integrate that on the RHS.
To do that, use the substitution \(\displaystyle u = \sinh^{-1}x\) so that \(\displaystyle x = \sinh(u)\) and then later write \(\displaystyle \cosh(u)=(e^{u}+e^{-u})/2\).
It's still messy after that (putting it back in terms of x).
Tiggsy
