Question

The graph shows the solution to the initial value problem y'(t)=mt, y(t0)= -4

Answer 1

$y'(t)=mt \\ \\ y=\int {mt} \, dt=m \int{t} \,dt=m \frac{t^2}{2} +C \\ \\y= \frac{mt^2}{2}+C$

Now find equation of the graph. It passes through the point (2,3), and intersects y-axis at -2.

$y=at^2+b \\ \\3=a\times 2^2-2 \\ \\3=4a-2 \\ \\5=4a \\ \\a= \frac{5}{4} \\ \\y=\frac{5}{4} t^2-2 \\ \\y= \frac{m}{2} t^2+C \\ \\ \frac{m}{2} = \frac{5}{4} \\ \\m= \frac{5}{2} \\ \\C=-2 \\ \\y(t_0)=-4 \\ \\-4= \frac{5}{4} t_0^2-2 \\ \\ -2= \frac{5}{4} t_0^2 \\ \\ t_0^2=- \frac{8}{5} \\ \\t_0=\pm \sqrt{- \frac{8}{5} }$