Partial Differential Equations Questions with Worked Solutions

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)