(1) Both equations in (a) and (b) are separable.
(a)
[tex]\dfrac xy y' = \dfrac{2y^2+1}{x+1} \implies \dfrac{\mathrm dy}{y(2y^2+1)} = \dfrac{\mathrm dx}{x(x+1)}[/tex]
Expand both sides into partial fractions.
[tex]\left(\dfrac1y - \dfrac{2y}{2y^2+1}\right)\,\mathrm dy = \left(\dfrac1x - \dfrac1{x+1}\right)\,\mathrm dx[/tex]
Integrate both sides:
[tex]\ln|y| - \dfrac12 \ln\left(2y^2+1\right) = \ln|x| - \ln|x+1| + C[/tex]
[tex]\ln\left|\dfrac y{\sqrt{2y^2+1}}\right| = \ln\left|\dfrac x{x+1}\right| + C[/tex]
[tex]\dfrac y{\sqrt{2y^2+1}} = \dfrac{Cx}{x+1}[/tex]
[tex]\boxed{\dfrac{y^2}{2y^2+1} = \dfrac{Cx^2}{(x+1)^2}}[/tex]
(You could solve for y explicitly, but that's just more work.)
(b)
[tex]e^{x+y}y' = 3x \implies e^y\,\mathrm dy = 3xe^{-x}\,\mathrm dx[/tex]
Integrate both sides:
[tex]e^y = -3e^{-x}(x+1) + C[/tex]
[tex]\ln(e^y) = \ln\left(C - 3e^{-x}(x+1)\right)[/tex]
[tex]\boxed{y = \ln\left(C - 3e^{-x}(x+1)\right)}[/tex]
(2)
(a)
[tex]y' + \sec(x)y = \cos(x)[/tex]
Multiply both sides by an integrating factor, sec(x) + tan(x) :
[tex](\sec(x)+\tan(x))y' + \sec(x) (\sec(x) + \tan(x)) y = \cos(x) (\sec(x) + \tan(x))[/tex]
[tex](\sec(x)+\tan(x))y' + (\sec^2(x) + \sec(x)\tan(x)) y = 1 + \sin(x)[/tex]
[tex]\bigg((\sec(x)+\tan(x))y\bigg)' = 1 + \sin(x)[/tex]
Integrate both sides and solve for y :
[tex](\sec(x)+\tan(x))y = x - \cos(x) + C[/tex]
[tex]y=\dfrac{x-\cos(x) + C}{\sec(x) + \tan(x)}[/tex]
[tex]\boxed{y=\dfrac{(x+C)\cos(x) - \cos^2(x)}{1+\sin(x)}}[/tex]
(b)
[tex]y' + y = \dfrac{e^x-e^{-x}}2[/tex]
(Note that the right side is also written as sinh(x).)
Multiply both sides by e ˣ :
[tex]e^x y' + e^x y = \dfrac{e^{2x}-1}2[/tex]
[tex]\left(e^xy\right)' = \dfrac{e^{2x}-1}2[/tex]
Integrate both sides and solve for y :
[tex]e^xy = \dfrac{e^{2x}-2x}4 + C[/tex]
[tex]\boxed{y=\dfrac{e^x-2xe^{-x}}4 + Ce^{-x}}[/tex]
(c) I've covered this in an earlier question of yours.
(d)
[tex]y'=\dfrac y{x+y}[/tex]
Multiply through the right side by x/x :
[tex]y' = \dfrac{\dfrac yx}{1+\dfrac yx}[/tex]
Substitute y(x) = x v(x), so that y' = xv' + v, and the DE becomes separable:
[tex]xv' + v = \dfrac{v}{1+v}[/tex]
[tex]xv' = -\dfrac{v^2}{1+v}[/tex]
[tex]\dfrac{1+v}{v^2}\,\mathrm dv = -\dfrac{\mathrm dx}x[/tex]
[tex]-\dfrac1v + \ln|v| = -\ln|x| + C[/tex]
[tex]\ln\left|\dfrac yx\right| -\dfrac xy = C - \ln|x|[/tex]
[tex]\ln|y| - \ln|x| -\dfrac xy = C - \ln|x|[/tex]
[tex]\boxed{\ln|y| -\dfrac xy = C}[/tex]
