differentials that I can't solve correctly I can't solve these differential, someone can help me with a step by step solution?  thanks
$$y'+ty=t^3$$ 
$$y'=3t^2y+4t^2$$
I tried the first integrating by $$e^{\int tdt}$$ using $$p(t)=t$$ and $$q(t)=t^3$$
so I have $$ye^{\int\ tdt} = {\int}t^3e^{\int\ tdt}$$
but then I don't know how to proceed
 A: Use the integrating factor method. See the answer to your earlier question Differential Equation $ty'= 3t^2-y$ solution incorrect 
$$\frac{dy}{dx}+P(x)y=Q(x) \Rightarrow I=e^{\int P(x)dx}$$
Then $$yI=\int IQdx +c$$
A: $$ \frac{dy}{dt} + ty = t^3$$
Using Integrating factor method here:
Integrating factor $e^{\int t dt}= e^{\frac{t^2}{2}}$
Multiply through by integrating factor
$$e^{\frac{t^2}{2}}\frac{dy}{dt} + tye^{\frac{t^2}{2}} = t^3e^{\frac{t^2}{2}} $$
By reverse chain rule:
$$ \frac{d}{dt} ye^{\frac{t^2}{2}} = t^3e^{\frac{t^2}{2}}$$
Integrating both sides wrt t:
$$ ye^{\frac{t^2}{2}} = \int  t^3e^{\frac{t^2}{2}}\ dt $$
To integrate RHS:
Let $u = \frac{t^2}{2}$, 
$$\frac{du}{dt} = 2t $$
$$\int  t^3e^{\frac{t^2}{2}} = \frac{1}{2} \int ue^{\frac{u}{2}} \ du $$
You can now integrate by parts to find this integral and solve for y:
$$ \int ue^{\frac{u}{2}} \ du$$
Let $v = u$, $\frac{dp}{du} = e^{\frac{u}{2}}  $
$\frac{dv}{du} = 1  $, $p = 2e^{\frac{u}{2}} $
$\int ue^{\frac{u}{2}} \ du = 2e^{\frac{u}{2}}(u-2) + c $
Can you finish it off now?
Please use the same technique to solve the second equation. To read more, see the following link
A: Notice, we have $$\frac{dy}{dt}+ty=t^3$$
$$I.F.=e^{\int tdt}=e^{t^2/2}$$ Hence, the solution is given as $$y(I.F.)=\int t^3(I.F.)dt+c$$
$$ye^{t^2/2}=\int t^3 e^{t^2/2}dt+c$$
Let $\frac{t^2}{2}=u\implies tdt=du$
$$ye^{t^2/2}=\int 2u e^{u}du+c$$
$$ye^{t^2/2}=2u\int e^{u}du-2\int e^udu+c$$
$$ye^{t^2/2}= 2e^{u}(u-1)+c$$
$$ye^{t^2/2}=2e^{t^2/2}\left(\frac{t^2}{2}-1\right)+c$$
Similarly, 
$$\frac{dy}{dt}=3t^2y+4t^2$$
$$\frac{dy}{dt}=t^2(3y+4)$$
$$\frac{dy}{3y+4}=t^2dt$$
$$\int \frac{dy}{3y+4}=\int t^2dt$$
$$\frac{1}{3}\ln|3y+4|=\frac{t^3}{3}+c$$
