Differential Equation $ty'= 3t^2-y$ solution incorrect

Question

$$ty'= 3t^2-y$$

I can't solve this equation, someone that can help me with a solution step by step? my result is $\dfrac ct+t^3$ but the correct solution is $\dfrac ct+t^2$

Answer 1

$$ t\frac{dy}{dt} + y = 3t^2$$

Using reverse chain rule:

$$ \frac{d}{dt}yt = 3t^2$$

Integrating both sides wrt t:

$$ yt = \frac{3t^3}{3} + c $$

$$ y(t) = \frac{t^3}{t} + \frac{c}{t} = t^2 + ct^{-1}$$

Where $c$ is a constant of integration

Answer 2

You can use the integrating factor method. First rearrange so you have $$y'+\frac 1ty=3t^2$$ The integrating factor is $$e^{\int\frac 1tdt}=t$$ Then your solution is given by $$yt=\int3t \times tdt+c$$ And then you get the answer...