dy/dx is the same as y' and d²y/dx² is the same as y".
4.
y"-3y'+2y=0 has the general solution Aeˣ+Be²ˣ:
Put y=Aeˣ+Be²ˣ, then y'=Aeˣ+2Be²ˣ, y"=Aeˣ+4Be²ˣ so y"-3y'+2y=
Aeˣ+4Be²ˣ-3(Aeˣ+2Be²ˣ)+2(Aeˣ+Be²ˣ)=0.
This is the reverse procedure to solving y"-3y'+2y=0, which is treated like a quadratic equation D²-3D+2=(D-1)(D-2). The zeroes are coefficients of x in the exponentials.
But now we have to see what we get as the result of applying the DE to ⅙e⁻ˣ.
If y=⅙e⁻ˣ, y'=-⅙e⁻ˣ, y"=⅙e⁻ˣ, so y"-3y'+2y=⅙e⁻ˣ+½e⁻ˣ+⅓e⁻ˣ=e⁻ˣ.
y"-3y'+2y=e⁻ˣ contains no arbitrary constants and has the solution Aeˣ+Be²ˣ+⅙e⁻ˣ.
5.
y²=4C(x+C)=4Cx+4C², 2yy'=4C and C=½yy', so we can substitute for C:
y²=2yy'(x+½yy')=2xyy'+y²y'²; y=2xy'+yy'², or y=y'(2x+yy').