Diff Equation ???????

y=((x+1)⁴/2)+5/2

y=((x+1)⁴/2)+(5/2)(x+1)²

y'=2(x+1)³

y'-2y/(x+1)=(x+1)³ (given eqn)

y'=2(x+1)³-2y/(x+1)=

My mistake............
Opsie got it...........

its a simple linear diff eqn..

int factor = e^{∫(-2/(x+1))dx} =(x+1)^-2

so multiplying on both sides n doing all those traditional works...
finally,

∫d[y/(x+1)²] = ∫(x+1)dx
=> [y/(x+1)²] = x²/2 +x +k

given f(0)=3...
so,

3/1 = 0 +0 + k
=> k=3

hence the function is [y/(x+1)²] = x²/2 +x +3

simplifying we get.......

y=((x+1)⁴/2)+(5/2)(x+1)²

Ok got my mistake.....

Ok..........