Verified answer

dy/dx = (2x-4)/y

y.dy = (2x-4).dx

∫y.dy = ∫(2x-4).dx

y²/2 = (2 x x²/2 – 4x) + c

y²/2 = (x²-4x)+c

y²=2(x²-4x)+C

Could it t be that you mean :-

$ y dy   = $ (2x  - 4) dx

y²/2   = x²  - 4x  +  C

4(a)

i) y'=(2x-4)/y=>yy'=2(x-2)

ii)yy'=2(x-2)

=>

ydy=2(x-2)dx

=>

y^2=2(x^2-4x)+C

=>

y=+/-sqr[2x(x-4)+C]

(C=the constant of integration)

4(b)

y"+2y'=3y

=>

y"+2y'-3y=0

=>

(D+3)(D-1)y=0

(where Dy=dy/dx)

Let (D-1)y=z, then(D+3)z=0

=>

dz/z=-3dx

=>

ln(z)=C1-3x)

=>

z=e^(C1-3x)

(D-1)y=e^(C1-3x)

=>

y'-y=e^(C1-3x)

(the integrating factor=e^-x)

=>

d[ye^(-x)]=e^(-x+C1-3x)dx

=>

d[ye^(-x)]=-e^(C1-4x)d(C1-4x)/4

=>

ye^(-x)=-e^(C1-4x)/4+C2

(After integrations; C1, C2 are constants of integration)

=>

y=C2e^x-e^(C1-3x)/4

is the general solution.

(dy/dx) = (2x-4)/y, ie., ydy = (2x-4)dx. Integration gives (1/2)y^2 = x^2-4x+c/2, ie., 

y^2 = 2x^2-8x+c. Then y = (+/-)(2x^2-8x+c)^(1//2).

4a:

dy/dx = (2x - 4)/yy dy = (2x - 4) dx∫y dy = ∫(2x - 4) dx½y² = x² - 4x + Cy² = 2(x² - 4x + C)4b:We can write equation 4b in a linear form:y'' + 2y' - 3y = 0Since it's linear and homogeneous (the right side is 0), we can use Euler's guess:y = e^(rt)y' = re^(rt)y'' = r²e^(rt)r²e^(rt) + 2re^(rt) - 3e^(rt) = 0r² + 2r - 3 = 0We can solve for r and plug into y = e^(rt)

4.(a)i

dy/dx = (2x - 4)/y

y dy = (2x - 4) dx

4(a)ii

y dy = (2x - 4) dx

∫ y dy = ∫ (2x - 4) dx

∫ y dy = ∫ (2x dx - 4 dx)

y²/2 = x² - 4x + C'

2x² - y² - 8x + C = 0

integral of 2x-4 is actually x²–4x+C

where did you get ln y?