This documents explains the way to solve the first derivative equation with initial condition by using integrating factor method.

1. 3rd April 2020
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Initial Value Problem (IVP)
Find the solution from first order linear differential equation
𝑦′
+ 2𝑦 = 2 − 𝑒−4𝑡
(1)
with initial condition 𝑦(0) = 1.
Answer:
For first order linear differential, the equation has to fulfill term:
𝑦′
+ 𝑝( 𝑡) = 𝑔(𝑡) (2)
We use integrating factor method to solve this equation. The integrating factor method
is:
𝜇( 𝑡) = 𝑒∫ 𝑝( 𝑡) 𝑑𝑡
(3)
To find the solution we use equation:
𝑦( 𝑡) =
1
𝜇( 𝑡)
∫ 𝜇( 𝑡) 𝑔(𝑡)𝑑𝑡 (4)
So, from equation (1), we can write
𝑝( 𝑡) = 2 (5)
𝑔( 𝑡) = 2 − 𝑒−4𝑡
(6)
Substitute equation (5) to equation (3)
𝜇( 𝑡) = 𝑒∫ 𝑝( 𝑡) 𝑑𝑡
𝜇( 𝑡) = 𝑒∫2 𝑑𝑡
𝜇( 𝑡) = 𝑒2𝑡
(7)
Substitute equation (6) and (7) to equation (4)
𝑦 =
1
𝑒2𝑡 ∫(𝑒2𝑡(2 − 𝑒−4𝑡))𝑑𝑡
𝑦 = 𝑒−2𝑡
∫ 2𝑒2𝑡
− 𝑒−2𝑡
𝑑𝑡
𝑦 = 𝑒−2𝑡
(𝑒2𝑡
− (−
1
2
𝑒−2𝑡
) + 𝐶)
𝑦 = 𝑒−2𝑡
(𝑒2𝑡
+
1
2
𝑒−2𝑡
+ 𝐶)
𝑦 = ( 𝑒−2𝑡
. 𝑒2𝑡) + (𝑒−2𝑡
.
1
2
𝑒−2𝑡
) + ( 𝐶𝑒−2𝑡)

2. 3rd April 2020
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𝑦 = 1 +
1
2
𝑒−4𝑡
+ 𝐶𝑒−2𝑡
(8)
So, the equation (8) is the solution of differential equation (1). We apply the initial
condition where 𝑦(0) = 1 to equation (8)
𝑦(0) = 1 +
1
2
𝑒−4(0)
+ 𝐶𝑒−2(0)
1 = 1 +
1
2
. 1 + 𝐶. 1
1 =
3
2
+ 𝐶
𝐶 = 1 −
3
2
= −
1
2
(9)
Substitute equation (9) to the equation (8) and we get the final solution
𝑦 = 1 +
1
2
𝑒−4𝑡
−
1
2
𝑒−2𝑡
(10)
Sources:
Zachary S Tseng (2008, 2012)
Paul Dawkins (2018)