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  3. University Of East London (UEL)
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  5. Calculus

Exam (elaborations)

Partial Differential Equations Questions with Worked Solutions
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This document will allow university Math students to practice partial 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 be able to answe...

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  • November 13, 2024
  • 34
  • 2024/2025
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Partial Differential Equations

1) A smooth function 𝑧 = 𝑧(𝑥, 𝑦) satisfies the partial differential equation
1 ∂𝑧 ∂𝑧 2
2 ( ∂𝑥 + ∂𝑦
) = 8(5𝑥 + 3𝑦) . Using the transformation equations 𝑢 = 5𝑥 + 3𝑦 and
5𝑧
𝑣 = 9𝑥 − 9𝑦, calculate the general solution of this partial differential equation.

𝑢 = 5𝑥 + 3𝑦

∂𝑢
∂𝑥
=5
∂𝑢
∂𝑦
=3


𝑣 = 9𝑥 − 9𝑦

∂𝑣
∂𝑥
=9
∂𝑣
∂𝑦
=− 9

∂𝑧 ∂𝑧 ∂𝑢 ∂𝑧 ∂𝑣
∂𝑥
= ∂𝑢 ∂𝑥
+ ∂𝑣 ∂𝑥
∂𝑧 ∂𝑧 ∂𝑧
∂𝑥
= ∂𝑢
x5 + ∂𝑣
x 9
∂𝑧 ∂𝑧 ∂𝑧
∂𝑥
= 5 ∂𝑢 + 9 ∂𝑣


∂𝑧 ∂𝑧 ∂𝑢 ∂𝑧 ∂𝑣
∂𝑦
= ∂𝑢 ∂𝑦
+ ∂𝑣 ∂𝑦
∂𝑧 ∂𝑧 ∂𝑧
∂𝑦
= ∂𝑢
x3 + ∂𝑣
x −9
∂𝑧 ∂𝑧 ∂𝑧
∂𝑦
= 3 ∂𝑢 − 9 ∂𝑣


∂𝑧 ∂𝑧 1 ∂𝑧 ∂𝑧 2
Substitute ∂𝑥
and ∂𝑦
into 2 ( ∂𝑥 + ∂𝑦
) = 8(5𝑥 + 3𝑦)
5𝑧


1 ∂𝑧 ∂𝑧 ∂𝑧 ∂𝑧 2
2 [(5 ∂𝑢
+9 ∂𝑣
) + (3 ∂𝑢
−9 ∂𝑣
)] = 8(5𝑥 + 3𝑦)
5𝑧
𝑢 = 5𝑥 + 3𝑦
1 ∂𝑧 2
2 [8 ∂𝑢
] = 8𝑢
5𝑧
8 ∂𝑧 2
2 ∂𝑢
= 8𝑢
5𝑧
8 2
2 ∂𝑧 = 8𝑢 ∂𝑢
5𝑧
8 2
∫ 2 ∂𝑧 = ∫8𝑢 ∂𝑢
5𝑧
8 −2 2
∫ 5𝑧 ∂𝑧 = ∫8𝑢 ∂𝑢
3
8 8𝑢
− 5𝑧
= 3
+ 𝑓(𝑣)
The constant of integration will be another variable involving 𝑣.

, 3
8 40𝑢
− 𝑧
= 3
+ 5𝑓(𝑣)
3
8 40𝑢
𝑧
=− 3
− 5𝑓(𝑣)
𝑧 1
8
= 40𝑢
3
− 3
−5𝑓(𝑣)
𝑧 1
8
=− 40𝑢
3

3
+5𝑓(𝑣)
8
𝑧 =− 3
40𝑢
3
+5𝑓(𝑣)
8
𝑧(𝑥, 𝑦) =− 3
40(5𝑥+3𝑦)
3
+5𝑓(9𝑥−9𝑦)
3
40(5𝑥+3𝑦)
𝑧(𝑥, 𝑦) =− 8 ÷ [ 3
+ 5𝑓(9𝑥 − 9𝑦)]
3
40(5𝑥+3𝑦) +15𝑓(9𝑥−9𝑦)
𝑧(𝑥, 𝑦) =− 8 ÷ 3
3
𝑧(𝑥, 𝑦) =− 8 x 3
40(5𝑥+3𝑦) +15𝑓(9𝑥−9𝑦)
24
𝑧(𝑥, 𝑦) =− 3
40(5𝑥+3𝑦) +15𝑓(9𝑥−9𝑦)




FINAL ANSWER:

24
𝑧(𝑥, 𝑦) =− 3
40(5𝑥+3𝑦) +15𝑓(9𝑥−9𝑦)

, 2) A smooth function 𝑧 = 𝑧(𝑥, 𝑦) satisfies the partial differential equation
1 ∂𝑧 ∂𝑧 2
2 ( ∂𝑥 + ∂𝑦
) = 2(3𝑥 + 8𝑦) . Using the transformations equations 𝑢 = 3𝑥 + 8𝑦
8𝑧
and 𝑣 = 17𝑥 − 17𝑦, calculate the general solution of this partial differential equation.

𝑢 = 3𝑥 + 8𝑦

∂𝑢
∂𝑥
=3
∂𝑢
∂𝑦
=8


𝑣 = 17𝑥 − 17𝑦

∂𝑣
∂𝑥
= 17
∂𝑣
∂𝑦
=− 17

∂𝑧 ∂𝑧 ∂𝑢 ∂𝑧 ∂𝑣
∂𝑥
= ∂𝑢 ∂𝑥
+ ∂𝑣 ∂𝑥
∂𝑧 ∂𝑧 ∂𝑧
∂𝑥
= ∂𝑢
x 3 + ∂𝑣 x 17
∂𝑧 ∂𝑧 ∂𝑧
∂𝑥
= 3 ∂𝑢 + 17 ∂𝑣

∂𝑧 ∂𝑧 ∂𝑢 ∂𝑧 ∂𝑣
∂𝑦
= ∂𝑢 ∂𝑦
+ ∂𝑣 ∂𝑦
∂𝑧 ∂𝑧 ∂𝑧
∂𝑦
= ∂𝑢
x 8 + ∂𝑣 x − 17
∂𝑧 ∂𝑧 ∂𝑧
∂𝑦
= 8 ∂𝑢 − 17 ∂𝑣

∂𝑧 ∂𝑧 1 ∂𝑧 ∂𝑧 2
Substitute ∂𝑥
and ∂𝑦
into 2 ( ∂𝑥 + ∂𝑦
) = 2(3𝑥 + 8𝑦)
8𝑧


1 ∂𝑧 ∂𝑧 ∂𝑧 ∂𝑧 2
2 [(3 ∂𝑢
+ 17 ∂𝑣
) + (8 ∂𝑢
− 17 ∂𝑣
)] = 2(3𝑥 + 8𝑦)
8𝑧
𝑢 = 3𝑥 + 8𝑦
1 ∂𝑧 2
2 [11 ∂𝑢
] = 2𝑢
8𝑧
11 ∂𝑧 2
2 ∂𝑢
= 2𝑢
8𝑧
11 2
2 ∂𝑧 = 2𝑢 ∂𝑢
8𝑧
11 2
∫ 2 ∂𝑧 = ∫2𝑢 ∂𝑢
8𝑧
11 −2 2
∫ 8
𝑧 ∂𝑧 = ∫2𝑢 ∂𝑢
3
11 2𝑢
− 8𝑧
= 3
+ 𝑓(𝑣)
The constant of integration will be another variable involving 𝑣.
3
11 16𝑢
− 𝑧
= 3
+ 8𝑓(𝑣)
3
11 16𝑢
𝑧
=− 3
− 8𝑓(𝑣)

, 𝑧 1
11
= 16𝑢
3
− 3
−8𝑓(𝑣)
𝑧 1
11
=− 16𝑢
3

3
+8𝑓(𝑣)
11
𝑧 =− 16𝑢
3

3
+8𝑓(𝑣)
11
𝑧(𝑥, 𝑦) =− 3
16(3𝑥+8𝑦)
3
+4𝑓(17𝑥−17𝑦)
3
16(3𝑥+8𝑦)
𝑧(𝑥, 𝑦) =− 11 ÷ [ 3
+ 4𝑓(17𝑥 − 17𝑦)]
3
16(3𝑥+8𝑦) +12𝑓(17𝑥−17𝑦)
𝑧(𝑥, 𝑦) =− 11 ÷ 3
3
𝑧(𝑥, 𝑦) =− 11 x 3
16(3𝑥+8𝑦) +12𝑓(17𝑥−17𝑦)
33
𝑧(𝑥, 𝑦) =− 3
16(3𝑥+8𝑦) +12𝑓(17𝑥−17𝑦)




FINAL ANSWER:

33
𝑧(𝑥, 𝑦) =− 3
16(3𝑥+8𝑦) +12𝑓(17𝑥−17𝑦)

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