**Calculus problem how do you do it?**

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*Solve the initial-value problem. dx/dt + 2tx = x , x(0)=5. Use your solution to compute x(3). Please show work and how do you do it? *

Rearrange to dx/dt = x – 2tx

dx/dt = x(1 – 2t)

split & integrate

Int 1/x dx = Int (1-2t) dt

ln|x| = t – t^2 + c

x(0)=5 ? are you actually saying that when t = 0, x = 5 (this makes more sense)

assuming this is the case:

ln5 = c

so ln|x| = t – t^2 + ln5

t = 3: ln|x| = 3 – 9 + ln5 = -6 + ln5

x = e^(-6 + ln5) = 0.012

**Initial Value Problem Example 1**