Find the specific solution of the differential equation.?

Question

1/2 dy/dt= (y^2+y)t (y>0)

Given that y(0)=2

Answer 1

Multiply through by 2dt, divide through by (y^2+y) to get:
dy/(y²+y) = 2t dt
The right side is just d(t²)
The left side is integrable by partial fractions:
1/(y²+y) = 1/y - 1/(y+1)

Specifically, integrate the left side from 2 to y
and integrate the right side from 0 to t:
ln (y/(y+1))|y - ln (y/(y+1))|2 = t²|t - t²|0
so:
ln (y/(y+1)) - ln(2/3) = t^2
Exponentiating:
y/(y+1) = (2/3) * e^t²
or:
(y+1-1)/(y+1) = (2/3) * e^t²
or:
1-1/(y+1) = (2/3) * e^t²
or:
1/(y+1) = 1-2e^t²/3
or:
y=1/(1-2e^t²/3) - 1
or:
y = 2e^t²/(3-2e^t²)

Note: I hope the browser renders it OK, but if not: t² = t*t = t squared.

Answer 2

Write it as
dy/(y^2+y)= 2tdt
and integrate to get
log[Cy/(1+y)] = t^2
i.e.,
Cy/(1+y) = exp(t^2)
Find C from y(0)=2 and you can solve for y(t).