Respuesta :

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]

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