y' + 2ty = t^3 where y(1) = 1?

Question

Find the solution of the following initial-value problem

Answer 1

xyz^t+x342-y^(8)= z

Answer 2

dy/dt+2t*y=0 so dy/y =-2t dt
ln IyI =-t^2+c
so
y= C*e^(-t^2)
Now find a particular solution of the equation with 2nd side
By inspection y = at^2+bt+c
y´=2at +b
2at+b +2at^3+2bt^2+2ct =t^3
2a=1 a=1/2
a+c= =0 c=-1/2 and b=0
y=1/2(t^2-1)
General solution
y=Ce^(-t^2)+1/2(t^2-1)
y(1) = 1=C/e so C=e and
y =e^(-t^2+1) +1/2(t^2-1)

Answer 3

You need an integrating factor.
I expect it is exp[t^2].

Use it to find the general solution, then feed in the initial condition.