Exam (elaborations)

First Order Ordinary Differential Equations Worksheet with Full Worked Solutions

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Course
Calculus

Institution
University Of East London (UEL)

This document will allow university Math students to practicing first order ordinary Differential equations. All the questions are similar in style and full worked solutions have been provided. This will allow you to understand how a certain question in this document is answered and then you will b...

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first order odes
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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)

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