The differential equation is separable.
What does it look like after we separate the variables?
[tex]dy=(3t^2+1)dt[/tex]
Let's integrate both sides of the equation.
[tex] \int\limits dy= \int\limits(3t^2+1)dt [/tex]
[tex]y=t^3+t+C[/tex]
What value of C satisfies the initial condition [tex]y(1)=5[/tex] ?
Let's substitute [tex]t=1[/tex] and [tex]y=5 [/tex] into the equation and solve for C.
[tex]5=1^3+1+C [/tex]
[tex]C=3[/tex]
Now use this value of C to find t when [tex]y=3[/tex]
[tex]3=t^3+t+3[/tex]
[tex]0=t^3+t[/tex]
[tex]0=(t^3+1)*t[/tex]
[tex]t=0[/tex]
