First Order Ordinary Differential Equations Worksheet with Full Worked Solutions

First Order Ordinary Differential Equations

𝑥+4𝑦 𝑑𝑦 2
1) 𝑒 𝑑𝑥
− (5 − 𝑥) = 0


Find a general solution to the above differential equation

𝑥 4𝑦 𝑑𝑦 2
𝑒 x𝑒 𝑑𝑥
= (5 − 𝑥)


We can now seperate the variables by putting all the 𝑥 terms on one side of the
equation and all the 𝑦 terms on the other.

2
4𝑦 (5−𝑥)
𝑒 𝑑𝑦 = 𝑥 𝑑𝑥
𝑒


2
4𝑦 (5−𝑥)
∫𝑒 𝑑𝑦 = ∫ 𝑥 𝑑𝑥
𝑒


4𝑦 −𝑥 2
∫𝑒 𝑑𝑦 = ∫𝑒 (5 − 𝑥) 𝑑𝑥

−𝑥 2
We can solve ∫𝑒 (5 − 𝑥) 𝑑𝑥 using integration by parts

𝑢𝑣 − ∫𝑣 𝑑𝑢

Choose 𝑢 to be a term that you can easily differentiate
Choose 𝑑𝑣 to be a term that you can easily integrate

2 −𝑥
𝑢 = (5 − 𝑥) 𝑑𝑣 = 𝑒 𝑑𝑥
−𝑥
𝑑𝑢 = 2(5 − 𝑥) x − 1 𝑣 =− 𝑒
𝑑𝑢 =− 2(5 − 𝑥)

−𝑥 2 −𝑥 2 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) − ∫ − 𝑒 (− 2(5 − 𝑥)) 𝑑𝑥
−𝑥 2 −𝑥
=− 𝑒 (5 − 𝑥) − ∫2𝑒 (5 − 𝑥) 𝑑𝑥
−𝑥 2 −𝑥
=− 𝑒 (5 − 𝑥) − 2∫𝑒 (5 − 𝑥) 𝑑𝑥

−𝑥
Calculate ∫𝑒 (5 − 𝑥) 𝑑𝑥 using integration by parts again

𝑢𝑣 − ∫𝑣 𝑑𝑢

, −𝑥
𝑢=5 −𝑥 𝑑𝑣 = 𝑒 𝑑𝑥
−𝑥
𝑑𝑢 =− 1 𝑑𝑥 𝑣 =− 𝑒
𝑑𝑢 =− 𝑑𝑥




−𝑥 −𝑥 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) − ∫ − 𝑒 (− 1) 𝑑𝑥
−𝑥 −𝑥 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) − ∫𝑒 𝑑𝑥
−𝑥 −𝑥 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) − − 𝑒
−𝑥 −𝑥 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) + 𝑒

−𝑥 2 −𝑥 2 −𝑥
∫𝑒 (5 − 𝑥) 𝑑𝑥 =− 𝑒 (5 − 𝑥) − 2∫𝑒 (5 − 𝑥) 𝑑𝑥
−𝑥 2 −𝑥 −𝑥
=− 𝑒 (5 − 𝑥) − 2(− 𝑒 (5 − 𝑥) + 𝑒 )
−𝑥 2 −𝑥 −𝑥
=− 𝑒 (5 − 𝑥) + 2𝑒 (5 − 𝑥) − 2𝑒 +𝐶

4𝑦 −𝑥 2
∫𝑒 𝑑𝑦 = ∫𝑒 (5 − 𝑥) 𝑑𝑥
1 4𝑦 −𝑥 2 −𝑥 −𝑥
4
𝑒 =− 𝑒 (5 − 𝑥) + 2𝑒 (5 − 𝑥) − 2𝑒 +𝐶
1 4𝑦 −𝑥 2
4
𝑒 = 𝑒 (− (5 − 𝑥) + 2(5 − 𝑥) − 2) + 𝐶
1 4𝑦 −𝑥 2
4
𝑒 = 𝑒 (− (𝑥 − 10𝑥 + 25) + 2(5 − 𝑥) − 2) + 𝐶
1 4𝑦 −𝑥 2
4
𝑒 = 𝑒 (− 𝑥 + 10𝑥 − 25 + 10 − 2𝑥 − 2) + 𝐶
1 4𝑦 −𝑥 2
4
𝑒 = 𝑒 (− 𝑥 + 8𝑥 − 17) + 𝐶
4𝑦 −𝑥 2
𝑒 = 4𝑒 (− 𝑥 + 8𝑥 − 17) + 4𝐶
Let 𝐾 = 4𝐶
4𝑦 −𝑥 2
𝑒 = 4𝑒 (− 𝑥 + 8𝑥 − 17) + 𝐾
−𝑥 2
4𝑦 = 𝑙𝑛(4𝑒 (− 𝑥 + 8𝑥 − 17) + 𝐾)
1 −𝑥 2
𝑦= 4
𝑙𝑛(4𝑒 (− 𝑥 + 8𝑥 − 17) + 𝐾)


FINAL ANSWER:

1 −𝑥 2
𝑦= 4
𝑙𝑛(4𝑒 (− 𝑥 + 8𝑥 − 17) + 𝐾)

, 2𝑥+7𝑦 𝑑𝑦 2
2) 𝑒 𝑑𝑥
− (8 − 𝑥) = 0


Find a general solution to the above differential equation

2𝑥 7𝑦 𝑑𝑦 2
𝑒 x𝑒 𝑑𝑥
= (8 − 𝑥)


We can now seperate the variables by putting all the 𝑥 terms on one side of the
equation and all the 𝑦 terms on the other.

2
7𝑦 (8−𝑥)
𝑒 𝑑𝑦 = 2𝑥 𝑑𝑥
𝑒


2
7𝑦 (8−𝑥)
∫𝑒 𝑑𝑦 = ∫ 2𝑥 𝑑𝑥
𝑒


7𝑦 −2𝑥 2
∫𝑒 𝑑𝑦 = ∫𝑒 (8 − 𝑥) 𝑑𝑥

−2𝑥 2
We can solve ∫𝑒 (8 − 𝑥) 𝑑𝑥 using integration by parts

𝑢𝑣 − ∫𝑣 𝑑𝑢

Choose 𝑢 to be a term that you can easily differentiate
Choose 𝑑𝑣 to be a term that you can easily integrate

2 −2𝑥
𝑢 = (8 − 𝑥) 𝑑𝑣 = 𝑒 𝑑𝑥
1 −2𝑥
𝑑𝑢 = 2(8 − 𝑥) x − 1 𝑑𝑥 𝑣 =− 2
𝑒
𝑑𝑢 =− 2(8 − 𝑥) 𝑑𝑥

−2𝑥 2 1 −2𝑥 2 1 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) − ∫ − 2
𝑒 (− 2(8 − 𝑥)) 𝑑𝑥
1 −2𝑥 2 −2𝑥
=− 2
𝑒 (8 − 𝑥) − ∫𝑒 (8 − 𝑥) 𝑑𝑥

−2𝑥
Calculate ∫𝑒 (8 − 𝑥) 𝑑𝑥 using integration by parts again

𝑢𝑣 − ∫𝑣 𝑑𝑢

1 −2𝑥
𝑢=8 −𝑥 𝑣 =− 2
𝑒
−2𝑥
𝑑𝑢 =− 1 𝑑𝑥 𝑑𝑣 = 𝑒 𝑑𝑥

, −2𝑥 1 −2𝑥 1 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) − ∫ − 2
𝑒 (− 1) 𝑑𝑥
−2𝑥 1 −2𝑥 1 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) − ∫ 2 𝑒 𝑑𝑥
−2𝑥 1 −2𝑥 1 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) − (− 4
𝑒 )
−2𝑥 1 −2𝑥 1 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) + 4
𝑒



−2𝑥 2 1 −2𝑥 2 −2𝑥
∫𝑒 (8 − 𝑥) 𝑑𝑥 =− 2
𝑒 (8 − 𝑥) − ∫𝑒 (8 − 𝑥) 𝑑𝑥
1 −2𝑥 2 1 −2𝑥 1 −2𝑥
=− 2
𝑒 (8 − 𝑥) − (− 2
𝑒 (8 − 𝑥) + 4
𝑒 )
1 −2𝑥 2 1 −2𝑥 1 −2𝑥
=− 2
𝑒 (8 − 𝑥) + 2
𝑒 (8 − 𝑥) − 4
𝑒 +𝐶

7𝑦 −2𝑥 2
∫𝑒 𝑑𝑦 = ∫𝑒 (8 − 𝑥) 𝑑𝑥
1 7𝑦 1 −2𝑥 2 1 −2𝑥 1 −2𝑥
7
𝑒 =− 2
𝑒 (8 − 𝑥) + 2
𝑒 (8 − 𝑥) − 4
𝑒 +𝐶
1 7𝑦 1 −2𝑥 2 1
7
𝑒 = 2
𝑒 (− (8 − 𝑥) + (8 − 𝑥) − 2
)+𝐶
1 7𝑦 1 −2𝑥 2 1
7
𝑒 = 2
𝑒 (− (𝑥 − 16𝑥 + 64) + (8 − 𝑥) − 2
)+ 𝐶
1 7𝑦 1 −2𝑥 2 1
7
𝑒 = 2
𝑒 (− 𝑥 + 16𝑥 − 64 + 8 − 𝑥 − 2
) +𝐶
1 7𝑦 1 −2𝑥 2 113
7
𝑒 = 2
𝑒 (− 𝑥 + 15𝑥 − 2 )+ 𝐶
7𝑦 7 −2𝑥 2 113
𝑒 = 2
𝑒 (− 𝑥 + 15𝑥 − 2 ) + 7𝐶
Let 𝐾 = 7𝐶
7𝑦 7 −2𝑥 2 113
𝑒 = 2
𝑒 (− 𝑥 + 15𝑥 − 2
) +𝐾
7 −2𝑥 2 113
7𝑦 = 𝑙𝑛( 2 𝑒 (− 𝑥 + 15𝑥 − 2
) + 𝐾)
1 7 −2𝑥 2 113
𝑦= 7
𝑙𝑛( 2 𝑒 (− 𝑥 + 15𝑥 − 2
) + 𝐾)




FINAL ANSWER:

1 7 −𝑥 2 113
𝑦= 7
𝑙𝑛( 2 𝑒 (− 𝑥 + 15𝑥 − 2
) + 𝐾)