initial value problem?

Question

(dy/dt)-(3/t)y = t^3 , y(1) = 4

Answer 1

This is a linear first order ODE, so we can find an integrating factor as e^(∫(-3/t) dt) = e^(-3 ln t) = t^-3. So we get
t^(-3) y' - 3t^(-4) y = 1
<=> d/dt (t^(-3) y) = 1
<=> t^-3 y = t + c
<=> y = t^4 + ct^3
Plug in the initial condition:
y(1) = 4 <=> 1^4 + c.1^3 = 4 => c = 3.
So the solution is
y = t^4 + 3t^3.

Answer 2

linear
P= -3/t

Q= t^3

I.F. = e^[integral P dt] = e^[-3lnt] = e^[ln(t^(-3))] = t^(-3)

solution is y*(I.F.) = integral {Q*(I.F.) dt} +c

y[ t^(-3) ] = integral {( t^(-3) )*( t^3)}dt +C

y[ t^(-3) ] = integral {1}dt +C

y[ t^(-3) ] = t + C

or y = t^4 + C ( t^3)

applying y(1) = 4

4 = 1 + C
C = 3

y = t^4 + 3 ( t^3)

Answer 3

what the heck does this q mean?