Question

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

find:
m=
t0=
y(t)=

Answer 1

x^2-4? or t^2 - 4?

Answer 2

t^2-4

Answer 3

why do you say "presumably" y=t^2-4 ???

Answer 4

there seems to be some missing info here, but you can think of this as\[y'(t)=mt\implies y(t)=\int mtdt=\frac m2t^2+y(t_0)\]which we can use to find m and \(y(t_0)\) if we know the graph