Solve linear differential equation

Question

So I have the following linear differential equation

$$t\frac{dy}{dt}-3y=t^4$$

My first step was to divide through by $t$ to give $$\frac{dy}{dt}-3t^{-1}y=t^3$$

Then to find the integrating factor $e^{\int-3t^{-1}dt}$ which gives $t^{-3}$

I know multiple each side by the integrating factor to get $$\frac{dy/dt}{t^3}-\frac{3y}{t^4}=1$$

Now i cant remember the next step, could anyone continue this problem for me?

Answer 1

So you're right in obtaining integrating factor of $t^{-3}$ and multiplying through by that. This gives the following:

$$ t^{-3}\frac{dy}{dt} - 3yt^{-4} = 1$$

$$ \frac{d}{dt}(yt^{-3}) = 1$$

Integrating through gives you:

$$ yt^{-3} = \int 1 \ dt $$

Can you finish it from here? Note that multiplying through by an integrating factor makes the differential equation exact which makes it easier to solve.

Answer 2

It is not difficult to realize that $y(t)=t^4$ is a solution, then by setting $y(t)=t^4+f(t)$ we are left with the differential equation:

$$ t\, f'(t) - 4\, f(t) = 0 $$ that is separable. That gives that every solution is of the form $y(t) = K\,t^4$.