12t1dydt 9y27t for t1 y05 solve the initial value problemSo

12(t+1)dy/dt - 9y=27t for t>-1, y(0)=5 solve the initial value problem

Divide all over by 12(t+1) :

dy/dt + (-3/4(t+1))*y = 9t/(4(t+1))

Integrating factor = e^(integral of (-3/4(t+1)*dt)

e^(-3/4*ln|t + 1|)

e^(ln|t + 1)|^(-3/4))

(t + 1)^(-3/4)

Multiply the DE by the untegrating factor :

(t + 1)^(-3/4) * [dy/dt + (-3/4(t+1))*y = 9t/(4(t+1))]

d/dt of (y * (t + 1)^(-3/4)) = 9t(t+1)^(-3/4) / (4(t+1))

Integrating both sides :

y * (t + 1)^(-3/4) = 3(3t + 4)/(t + 1)^(3/4) + C

y = 3(3t + 4) + C(t + 1)^(3/4)

y = C(t + 1)^(3/4) + 9t + 12

y(0) = 5 :

5 = C + 9(0) + 12

5 = C + 12

C = -7

So, solution, y = C(t + 1)^(3/4) + 9t + 12
becomes

y = -7(t + 1)^(3/4) + 9t + 12