Question

Question is in the attachment

Answer 1

The differential equation is separable.

What does it look like after we separate the variables?


`dy=(3t^2+1)dt`

Let's integrate both sides of the equation.


`\int\limits dy= \int\limits(3t^2+1)dt`


`y=t^3+t+C`

What value of C satisfies the initial condition
`y(1)=5` ?

Let's substitute
`t=1` and
`y=5` into the equation and solve for C.


`5=1^3+1+C`


`C=3`

Now use this value of C to find t when
`y=3`


`3=t^3+t+3`


`0=t^3+t`


`0=(t^3+1)*t`


`t=0`