# Thread: Differential Equation - quick question

1. ## Differential Equation - quick question

The problem is: Find the continuous solution to the differential equation
dy/dt = 2t+2 if t=<2, or 3t if 2=<t
initial condition y(0) = 2

using that initial condition and the first equation, i solved and got that y(t) = t^2 + 2t + 2
i know that function is continuous, but does it also apply to the 3t or do i need to solve a separate equation? if so, how would i go about doing this?

2. Originally Posted by mistykz
The problem is: Find the continuous solution to the differential equation
dy/dt = 2t+2 if t=<2, or 3t if 2=<t
initial condition y(0) = 2

using that initial condition and the first equation, i solved and got that y(t) = t^2 + 2t + 2
i know that function is continuous, but does it also apply to the 3t or do i need to solve a separate equation? if so, how would i go about doing this?
Use the solution you have to find the solution value at t=2, then use that as the initial condition to solve the second equation for t>=2.

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3. Alright, I figured it out. Thank you!