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Prof. N. R. Kharat
Engineering Mathematics I
UNIT III: Partial Differentiation and Applications
Partial Differentiation:
Given a function of two variables, ƒ( x, y) the derivative with respect to x only
(treating y as a constant) is called the partial derivative of ƒ with respect to x
It is denoted by either
𝜕𝑓
𝜕𝑥
𝑜𝑟 𝑓𝑥
Similarly, the derivative of ƒ with respect to y only (treating x as a constant) is
called the partial derivative of ƒ with respect to y. It is denoted by either
𝜕𝑓
𝜕𝑦
or fy
Differentiate ƒ with respect to x twice. (i.e differentiate ƒ with respect to x; then
differentiate the result with respect to x again.)
∂f
∂2 f
∂x ∂x
∂x2
∂
( )=
𝑜𝑟 𝑓𝑥𝑥
Differentiate ƒ with respect to y twice. (i.e. differentiate ƒ with respect to y; then
differentiate the result with respect to y again.)
∂f
∂2 f
∂y ∂y
∂y2
∂
( )=
𝑜𝑟 𝑓𝑦𝑦
Mixed partials derivative:

First differentiate ƒ with respect to x; then differentiate the result with respect
to y.
∂
∂f
( )=
∂y ∂x

∂2 f
∂y ∂x
𝑜𝑟 𝑓𝑥𝑦
First differentiate ƒ with respect to y; then differentiate the result with respect
to x.
∂
∂f
( )=
∂x ∂y
∂2 f
∂y ∂x
𝑜𝑟 𝑓𝑦𝑥
***************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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Prof. N. R. Kharat
Engineering Mathematics I
Example 1: If 𝑓(𝑥, 𝑦) = 3𝑥 2 𝑦 + 5𝑥 − 2𝑦 2 + 1 find 𝑓𝑥 𝑓𝑦 𝑓𝑥𝑥 𝑓𝑦𝑦 𝑓𝑥𝑦 𝑓𝑦𝑥
Solution: Let 𝑓(𝑥, 𝑦) = 3𝑥 2𝑦 + 5𝑥 − 2𝑦 2 + 1
𝑓𝑥 means differentiate 𝑓 with respect to x (treating y as a constant)
∂f
∂
= (3𝑥 2 𝑦 + 5𝑥 − 2𝑦 2 + 1 )
∂x ∂x
∂f
𝑓𝑥 =
= 6𝑥𝑦 + 5 − 0 + 0 = 6𝑥𝑦 + 5
∂x
𝑓𝑥 =
𝑓𝑦 𝑚𝑒𝑎𝑛𝑠 differentiate f with respect to y (treating x as a constant)
𝑓𝑦 =
∂f
∂
= (3𝑥 2 𝑦 + 5𝑥 − 2𝑦 2 + 1 )
∂y ∂y
𝑓𝑦 =
∂f
= 3𝑥 2 + 0 − 4𝑦 + 0 = 3𝑥 2 − 4𝑦
∂y
∂2 f
𝑜𝑟 𝑓𝑥𝑥 Differentiate ƒ with respect to x twice.
∂x2
∂2 f
𝑜𝑟 𝑓𝑥𝑥 =
∂x2
∂2 f
∂
∂f
∂
∂x ∂x
∂x
( )=
( 6𝑥𝑦 + 5) = 6𝑦
𝑜𝑟 𝑓𝑦𝑦 Differentiate ƒ with respect to y twice.
∂y2
∂2 f
𝑜𝑟 𝑓𝑦𝑦 =
∂y2
∂
∂f
( )=
∂y ∂y
∂
∂y
( 3𝑥 2 − 4𝑦) = −4
First differentiate ƒ with respect to x; then differentiate the result with respect
to y.
∂2 f
∂y ∂x
𝑜𝑟 𝑓𝑥𝑦 =
∂
∂f
∂
∂y ∂x
∂y
( )=
(6𝑥𝑦 + 5) = 6𝑥
First differentiate ƒ with respect to y; then differentiate the result with respect
to x.
∂2 f
∂x ∂y
𝑜𝑟 𝑓𝑦𝑥 =
∂
∂f
∂
∂x ∂y
∂x
( )=
(3𝑥 2 − 4𝑦) = 6𝑥
*************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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Prof. N. R. Kharat
Engineering Mathematics I
3
Example 2: If 𝑧 = tan(𝑦 + 𝑎𝑥) + (𝑦 − 𝑎𝑥 )2 then find the value of
∂2 z
∂2 z
∂x
∂y2
− 𝑎2
2
3
Solution: Let 𝑧 = tan(𝑦 + 𝑎𝑥) + (𝑦 − 𝑎𝑥 )2
∂2 z
To find 2 Differentiate 𝑧 with respect to 𝑥 twice
∂x
𝑧 = tan(𝑦 + 𝑎𝑥) + (𝑦
3
− 𝑎𝑥 )2
Differentiating 𝑧 with respect to 𝑥 partially treating y as constant
3
∂z
∂
𝐝
𝒅 𝒏
[tan(𝑦 + 𝑎𝑥 ) + (𝑦 − 𝑎𝑥 )2 ] ∵
=
𝐭𝐚𝐧 𝐱 = 𝐬𝐞𝐜𝟐 𝒙
𝒙 = 𝒏𝒙𝒏−𝟏
∂x ∂x
𝐝𝐱
𝒅𝒙
3
∂z
∂
3
∂
= sec 2(𝑦 + 𝑎𝑥 ) (𝑦 + 𝑎𝑥) + (𝑦 − 𝑎𝑥 )2−1 (𝑦 − 𝑎𝑥)
∂x
∂x
2
∂x
1
∂z
3
= sec 2(𝑦 + 𝑎𝑥 ) (0 + 𝑎) + (𝑦 − 𝑎𝑥 )2 (0 − 𝑎)
∂x
2
1
∂z
3𝑎
2(
(𝑦 − 𝑎𝑥 )2 … … … … (1)
= 𝑎 sec 𝑦 + 𝑎𝑥 ) −
∂x
2
Differentiating
∂z
∂x
with respect to 𝑥 partially treating y as constant
1
∂ ∂z
∂
3𝑎
(𝑦 − 𝑎𝑥 )2 ]
( ) = [𝑎 sec 2(𝑦 + 𝑎𝑥 ) −
∂x ∂x
∂x
2
1
∂2 z
∂
3𝑎 1
∂
2−1
(
)
(
)]
(
)
[
=
2
asec
𝑦
+
𝑎𝑥
sec
𝑦
+
𝑎𝑥
−
𝑦
−
𝑎𝑥
(𝑦 − 𝑎𝑥)
∂x2
∂x
2 2
∂x
∂2 z
∂
(
)
(
)
(
)
(𝑦 + 𝑎𝑥 )
=
2
asec
𝑦
+
𝑎𝑥
sec
𝑦
+
𝑎𝑥
tan
𝑦
+
𝑎𝑥
∂x2
∂x
1
3𝑎
(𝑦 − 𝑎𝑥 )−2 (−𝑎)
−
4
1
∂2 z
3𝑎
−2
2(
)
(
)
(
)
(
)
= 2 a sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥 𝑎 −
𝑦 − 𝑎𝑥 (−𝑎)
∂x2
4
1
∂2 z
3𝑎 2
−2
2
2(
)
(
)
(
)
= 2 a sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥 +
𝑦 − 𝑎𝑥
… … … … . . (𝐴)
∂x2
4
∂2 z
To find 2 Differentiate z with respect to y twice
∂y
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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Prof. N. R. Kharat
Engineering Mathematics I
3
𝑧 = tan(𝑦 + 𝑎𝑥) + (𝑦 − 𝑎𝑥 )2
Differentiating 𝑧 with respect to 𝑦 partially treating x as constant
3
∂z
∂
[tan(𝑦 + 𝑎𝑥) + (𝑦 − 𝑎𝑥 )2 ]
=
∂y ∂y
3
∂z
∂
3
∂
2(
)
(
)
= sec 𝑦 + 𝑎𝑥
(𝑦 + 𝑎𝑥) + 𝑦 − 𝑎𝑥 2−1 (𝑦 − 𝑎𝑥)
∂y
∂y
2
∂y
1
∂z
3
= sec 2(𝑦 + 𝑎𝑥 ) (1 + 0) + (𝑦 − 𝑎𝑥 )2 (1 − 0)
∂y
2
1
∂z
3
2(
)
(
)
= sec 𝑦 + 𝑎𝑥 + 𝑦 − 𝑎𝑥 2 … … … … (2)
∂y
2
Differentiating
∂z
with respect to 𝑦 partially treating x as constant
∂y
1
∂ ∂z
∂
3
( ) = [sec 2(𝑦 + 𝑎𝑥 ) + (𝑦 − 𝑎𝑥 )2]
∂y ∂y
∂y
2
1
∂2 z
∂
31
∂
2−1
(
)
(
)]
(
)
[
=
2
sec
𝑦
+
𝑎𝑥
sec
𝑦
+
𝑎𝑥
+
𝑦
−
𝑎𝑥
(𝑦 − 𝑎𝑥)
∂y 2
∂x
22
∂x
1
∂2 z
∂
3
−
(
)
(
)
(
)
(
)
(
)
= 2 sec 𝑦 + 𝑎𝑥 sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥
𝑦 + 𝑎𝑥 + 𝑦 − 𝑎𝑥 2 (1)
∂y 2
∂x
4
1
∂2 z
3𝑎
−2
2(
)
(
)
(
)
(
)
= 2 sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥 𝑎 +
𝑦 − 𝑎𝑥
… … … … . (3)
∂y 2
4
Multiplying by −a2
1
∂2 z
3𝑎 2
−2
2
2(
)
(
)
(
)
−a
=
−2
a
sec
𝑦
+
𝑎𝑥
tan
𝑦
+
𝑎𝑥
−
𝑦
−
𝑎𝑥
… … … … . . (𝐵)
∂y 2
4
2
Equation (A) + (B)
1
∂2 z
∂2 z
3𝑎 2
−2
2
2
2(
)
(
)
(
)
−a
= 2 a sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥 +
𝑦 − 𝑎𝑥
∂x2
∂y 2
4
1
3𝑎 2
−2
)
(
)
(
)
= −2 a sec 𝑦 + 𝑎𝑥 tan 𝑦 + 𝑎𝑥 −
𝑦 − 𝑎𝑥
4
2
2(
∂2 z
∂2 z
2
∴ 2−a
=0
∂x
∂y 2
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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Prof. N. R. Kharat
Engineering Mathematics I
Example 3: If 𝑢 = log(𝑥 3 + 𝑦 3 − 𝑥 2 𝑦 − 𝑥𝑦 2 ) then prove that
𝜕
𝜕 2
4
( + ) 𝑢=−
(𝑥 + 𝑦)2
𝜕𝑥 𝜕𝑦
Solution: Let 𝑢 = log(𝑥 3 + 𝑦 3 − 𝑥 2 𝑦 − 𝑥𝑦 2 )
𝑢 = log(𝑥 2 (𝑥 − 𝑦) − 𝑦 2 (𝑥 − 𝑦))
𝑢 = log[(𝑥 − 𝑦) (𝑥 2 − 𝑦 2 )]
𝑢 = log[(𝑥 − 𝑦) (𝑥 − 𝑦)(𝑥 + 𝑦)]
∵ (𝒂𝟐 − 𝒃𝟐 ) = (𝒂 + 𝒃)(𝒂 − 𝒃)
𝑢 = log[(𝑥 − 𝑦)2 (𝑥 + 𝑦)]
𝑢 = log(𝑥 − 𝑦)2 + log(𝑥 + 𝑦)
∵ 𝒍𝒐𝒈 𝒂𝒃 = 𝒍𝒐𝒈 𝒂 + 𝒍𝒐𝒈 𝒃
𝑢 = 2 log(𝑥 − 𝑦) + log(𝑥 + 𝑦)
∵ 𝐥𝐨𝐠 𝒂𝒏 = 𝒏 𝐥𝐨𝐠 𝒂
𝜕
𝜕 2
∂2
∂2
∂2
∂2 u
∂2 u
∂2 u
( + ) 𝑢 =( 2 +2
) u= 2+ 2
+
+
𝜕𝑥 𝜕𝑦
∂x
∂x ∂y ∂y 2
∂x
∂x ∂y
∂y 2
… … (A)
∂2 u
To find 2 means Differentiate 𝑢 with respect to 𝑥 twice
∂x
𝑢 = 2 log(𝑥 − 𝑦) + log(𝑥 + 𝑦)
Differentiating 𝑢 with respect to 𝑥 partially y is constant
∂u
∂
𝒅
𝟏
[2 log(𝑥 − 𝑦) + log(𝑥 + 𝑦) ]
=
∵
𝐥𝐨𝐠 𝒙 =
∂x ∂x
𝒅𝒙
𝒙
∂u
2
∂
1
∂
(𝑥 − 𝑦) +
(𝑥 + 𝑦)
=
(𝑥 − 𝑦) ∂x
(𝑥 + 𝑦) ∂x
∂x
∂u
2
1
(1 − 0) +
(1 + 0)
=
(𝑥 − 𝑦)
(𝑥 + 𝑦)
∂x
∂u
2
1
=
+
… … … … . (1)
(𝑥 − 𝑦)
(𝑥 + 𝑦)
∂x
Differentiating
∂u
with respect to 𝑥 partially
∂x
∂2 u
∂ ∂u
∂
2
1
(
)
[
]
=
=
+
(𝑥 + 𝑦)
∂x2
∂x ∂x
∂x (𝑥 − 𝑦)
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
∂2 u
−2
∂
−1
∂
(
)
(𝑥 + 𝑦)]
[
=
𝑥
−
𝑦
+
(𝑥 − 𝑦)2 ∂x
(𝑥 + 𝑦)2 ∂x
∂x2
∵
𝒅 𝟏
−𝟏
( )= 𝟐
𝒅𝒙 𝒙
𝒙
∂2 u
−2
−1
(
)
(1 + 0)]
[
=
1
−
0
+
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂x2
∂2 u
2
1
=
−
−
… … … … … … . . (2)
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂x2
∂2 u
To find 2 Differentiate 𝑢 with respect to 𝑦 twice
∂y
𝑢 = 2 log(𝑥 − 𝑦) + log(𝑥 + 𝑦)
Differentiating 𝑢 with respect to 𝑦 partially
∂u
∂
[2 log(𝑥 − 𝑦) + log(𝑥 + 𝑦) ]
=
∂y ∂y
∂u
2
∂
1
∂
(𝑥 − 𝑦) +
(𝑥 + 𝑦)
=
(𝑥 − 𝑦) ∂y
(𝑥 + 𝑦) ∂y
∂y
∂u
2
1
(0 − 1) +
(0 + 1)
=
(𝑥 − 𝑦)
(𝑥 + 𝑦)
∂y
∂u
−2
1
=
+
… … … … . (3)
(𝑥 − 𝑦)
(𝑥 + 𝑦)
∂y
Differentiating
∂u
with respect to 𝑦 partially
∂y
∂2 u
∂ ∂u
∂
−2
1
(
)
[
]
=
=
+
(𝑥 + 𝑦)
∂y 2
∂y ∂y
∂ (𝑥 − 𝑦)
∂2 u
−(−2) ∂
−1
∂
(
)
(𝑥 + 𝑦)]
[
=
𝑥
−
𝑦
+
(𝑥 − 𝑦)2 ∂y
(𝑥 + 𝑦)2 ∂y
∂y 2
∂2 u
2
−1
(
)
(0 + 1)]
[
=
0
−
1
+
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂y 2
∂2 u
2
1
=
−
… … … … … … . . (4)
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂y 2
∂2 u
Differentiating 𝑢 with respect to 𝑦 partially then 𝑥
∂x ∂y
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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Prof. N. R. Kharat
Engineering Mathematics I
∂2 u
∂ ∂u
∂
−2
1
)
= ( )= (
+
(𝑥 + 𝑦)
∂x ∂y ∂x ∂y
∂x (𝑥 − 𝑦)
𝑓𝑟𝑜𝑚 (3)
∂2 u
−(−2) ∂
−1
∂
=
(𝑥
−
𝑦)
+
(𝑥 + 𝑦)
(𝑥 − 𝑦)2 ∂x
(𝑥 + 𝑦)2 ∂x
∂x ∂y
∂2 u
2
−1
(
)
=
1
−
0
+
(1 + 0)
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂x ∂y
∂2 u
2
1
=
−
… … … … . . (5)
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
∂x ∂y
𝜕
𝜕 2
∂2 u
∂2 u
∂2 u
( + ) 𝑢=
+ 2
+
𝜕𝑥 𝜕𝑦
∂x2
∂x ∂y
∂y 2
From (2) (4) and (5)
𝜕
𝜕 2
2
1
2
1
2
1
( + ) 𝑢= −
−
+
2
[
−
]
−
−
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
𝜕𝑥 𝜕𝑦
= −
2
1
4
2
2
1
−
+
−
−
−
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2
(𝑥 − 𝑦)2
(𝑥 + 𝑦)2 (𝑥 − 𝑦)2
(𝑥 + 𝑦)2
=−
4
(𝑥 + 𝑦)2
***************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
𝐓𝐲𝐩𝐞 𝐈𝐈
Engineering Mathematics I
∂2 u
∂2 u
=
∂x ∂y ∂y ∂x
If f(x, y) is Homogeneous Function then
Example 1: If 𝑢 = log(𝑥 2 + 𝑦 2 ) verify
∂2 f
∂x ∂y
∂2 u
∂x ∂y
=
=
∂2 f
∂y ∂x
∂2 u
∂y ∂x
Solution: Let 𝑢 = log(𝑥 2 + 𝑦 2 )
∂u
Differentiating 𝑢 with respect to 𝑥 partially y is constant
∂x
∂u
∂
= log(𝑥 2 + 𝑦 2 )
∂x ∂x
∂u
1
∂
𝒅
𝟏
2
2)
(
= 2
𝑥
+
𝑦
∵
𝐥𝐨𝐠
𝒙
=
∂x 𝑥 + 𝑦 2 ∂x
𝒅𝒙
𝒙
∂u
1
2𝑥
(
)
= 2
2𝑥
+
0
=
… … … … . . (1)
(𝑥 2 + 𝑦 2 )
∂x 𝑥 + 𝑦 2
∂2 u
∂u
Differentiating
with respect to 𝑦 partially x is constant
∂y ∂x
∂x
∂2 u
∂ ∂u
∂
2𝑥
( 2
)
= ( )=
∂y ∂x ∂y ∂x
∂y 𝑥 + 𝑦 2
∂2 u
∂
1
)
= 2𝑥 ( 2
∂y ∂x
∂y 𝑥 + 𝑦 2
∵
𝒅 𝟏
𝟏
( )= − 𝟐
𝒅𝒙 𝒙
𝒙
∂2 u
2𝑥 (−1) ∂
(𝑥 2 + 𝑦 2 )
=
2
2
2
(𝑥 + 𝑦 ) ∂y
∂y ∂x
∂2 u
− 2𝑥
(0 + 2𝑦)
=
(𝑥 2 + 𝑦 2 )2
∂y ∂x
∂2 u
−4𝑥𝑦
=
… … … . . (𝐴)
(𝑥 2 + 𝑦 2 )2
∂y ∂x
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
𝑢 = log(𝑥 2 + 𝑦 2 )
∂u
Differentiating 𝑢 with respect to 𝑦 partially
∂y
∂u
∂
= log(𝑥 2 + 𝑦 2 )
∂y ∂y
∂u
1
∂
(𝑥 2 + 𝑦 2 )
= 2
2
∂y 𝑥 + 𝑦 ∂y
∂u
1
2𝑦
(
)
= 2
0
+
2𝑦
=
… … … … . . (2)
(𝑥 2 + 𝑦 2 )
∂y 𝑥 + 𝑦 2
∂2 u
∂u
Differentiating
with respect to 𝑥 partially constant y
∂x ∂y
∂y
∂2 u
∂ ∂u
∂
2𝑦
∂
1
( 2
)
(
)
= ( )=
=
2𝑦
∂x ∂y ∂x ∂y
∂x 𝑥 + 𝑦 2
∂x 𝑥 2 + 𝑦 2
∂2 u
2𝑦 (−1) ∂
(𝑥 2 + 𝑦 2 )
=
2
2
2
(
)
∂x ∂y
𝑥 +𝑦
∂x
∂2 u
− 2𝑦
(2𝑥 + 0)
=
(𝑥 2 + 𝑦 2 )2
∂y ∂x
∂2 u
−4𝑥𝑦
=
… … … . . (𝐵)
(𝑥 2 + 𝑦 2 )2
∂y ∂x
From (A) and (B)
∂2 u
∂2 u
=
∂x ∂y ∂y ∂x
***************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Variable to be treated as Constant:
Let f be function of x and y i.e. 𝑓(𝑥, 𝑦)
To find partial derivative with respect to x we have to treat y as constant
𝜕𝑓
( )
𝜕𝑥 𝑦
𝜕𝑓
( )
To find partial derivative with respect to y we have to treat x as constant
𝜕𝑦 𝑥
For Example:
𝜕𝑧
( )
Means 𝑧 is function of 𝑟, 𝜃
𝜕𝑟 𝜃
z is differentiable w. r .t 𝑟, 𝜃
here z is partially differentiated with respect to r and 𝜃 as constant.
𝜕𝑦
( ) Means y is function of 𝑟, 𝜃
y is differentiable w. r. t 𝑟, 𝜃
𝜕𝜃 𝑟
here y is partially differentiated with respect to 𝜃 and r as constant.
********
Example 1: If u = 2x + 3y , v = 3x − 2y
𝜕𝑢
𝜕𝑥
𝜕𝑦
𝜕𝑣
𝜕𝑥 𝑦
𝜕𝑢 𝑣
𝜕𝑣 𝑥 𝜕𝑦 𝑢
Find ( ) ( ) ( ) ( )
Solution:
u = 2x + 3y
… … … (1)
v = 3x − 2y
… … … (2)
𝑢, 𝑣 are functions of 𝑥, 𝑦 variable
𝜕𝑢
( )
𝜕𝑥 𝑦
𝑖. 𝑒. 𝑢, 𝑣 = 𝑓(𝑥, 𝑦)
𝑢 is function of 𝑥, 𝑦 variable
𝑖. 𝑒 𝑢 = 2𝑥 + 3𝑦
Diff u partially w. r. t. x keep y as constant
𝜕𝑢
… … … … … … . (𝐴)
( ) =2
𝜕𝑥 𝑦
𝜕𝑥
( )
𝜕𝑢 𝑣
𝑥 is function of 𝑢, 𝑣 variable
u = 2x + 3y
v = 3x − 2y
u = 2x + 3y
2y = 3x − v
u = 2x + 3y
y=
u = 2x + 3 (
Eliminate 𝑦
3x − v
2
3x − v
2
)
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
2u = 4x + 3(3x − v)
2u = 4x + 9x − 3v
2u + 3v = 13x
x=
𝜕𝑥
( )
𝜕𝑢 𝑣
2u + 3v
13
Diff partially w. r. t. u
𝜕𝑥
2
𝜕𝑢 𝑣
13
( ) =
keep v as constant
… … … … (𝐵)
𝜕𝑦
( )
𝑦 is function of 𝑣, 𝑥 variable
𝜕𝑣 𝑥
v = 3x − 2y
From (2)
⇒ 2y = 3x − v
⇒
y=
3x − v
2
Diff partially w. r. t. v keep x as constant
𝜕𝑦
( )
𝜕𝑣 𝑥
=
−1
… … … … … … . (𝐶 )
2
𝜕𝑣
( )
𝑣 is function of 𝑦, 𝑢 variable
𝜕𝑦 𝑢
u = 2x + 3y
v = 3x − 2y
u = 2x + 3y
v + 2y = 3x
u = 2x + 3y
x=
u = 2(
Eliminate 𝑥
v + 2y
3
v + 2y
) + 3y
3
3u = 2v + 4y + 9y
3u = 2v + 13y
2v = 3u − 13y
v=
3u – 13y
2
Diff partially w. r. t. y keep u as constant
𝜕𝑣
13
𝜕𝑦 𝑢
2
( ) =−
… … … … (𝐷)
From (A), (B), (C) and (D)
𝜕𝑢
𝜕𝑥
𝜕𝑦
𝜕𝑣
𝜕𝑥 𝑦
𝜕𝑢 𝑣
𝜕𝑣 𝑥 𝜕𝑦 𝑢
2
1
13
13
2
2
∴ ( ) ( ) ( ) ( ) = 2 ( ) (− ) (− ) = 1
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
*******************
Example 2: If x =
cos θ
r
sin θ
, y=
r
𝜕𝑥
𝜕𝑟
𝜕𝑦
𝜕𝑟
Find ( ) ( ) + ( ) ( )
𝜕𝑟 θ 𝜕𝑥 𝑦
𝜕𝑟 θ 𝜕𝑦 𝑥
Solution:
x=
y=
cos θ
… … … (1)
r
sin θ
… … … (2)
r
𝑥, 𝑦 are function 𝑜𝑓 𝑟, 𝜃 variable 𝑖. 𝑒. 𝑥, 𝑦 = 𝑓(𝑟, 𝜃)
𝜕𝑥
( )
𝑥 is function of 𝑟, 𝜃 variable
𝜕𝑟 θ
𝑖. 𝑒
x=
cos θ
r
Diff x partially w. r. t. r keep as 𝜃 constant
𝜕𝑥
cos θ
( ) =−
r2
𝜕𝑟 θ
𝜕𝑟
( )
𝜕𝑥 𝑦
cos θ
x=
x2 =
… … … … … … . (𝐴)
𝒅
𝟏
𝟏
𝒅𝒙 𝒙
𝒙𝟐
( )=−
r is function of 𝑥, 𝑦 variable
y=
r
cos2 θ
sin θ
r
y2 =
r2
Eliminate θ
sin2 θ
r2
squaring
adding
2
2
x + y =
cos2 θ
r2
r2 =
r2
cos2 θ + sin2 θ
x2 + y 2 =
x2 + y 2 =
+
sin2 θ
r2
1
r2
1
x2 + y2
r 2 = (x2 + y 2 )−1
taking square root on both side
r = (x2 + y 2 )−1/2
−3 𝜕
𝜕𝑟
1 2
2
(
)
(x2 + y 2 )
( ) =− x + y 2
𝜕𝑥 𝑦
2
𝜕𝑥
𝒅 𝒏
𝑑𝑥
𝒙 = 𝒏𝒙𝒏−𝟏
𝒅𝒙
𝑑𝑥
−3
𝜕𝑟
1 2
2) 2
(
( ) =− x + y
(2𝑥 + 0)
𝜕𝑥 𝑦
2
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
−3
1 2 cos θ
)
= (−1) ( 2 ) (
r
r
1 cos θ
)
= (−1) −3 (
r
r
cos θ
= − −2
r
= −r 2 cos θ
𝜕𝑦
( )
𝐱𝟐 + 𝐲𝟐 =
… … … (B)
𝑦 is function of 𝑟, 𝜃 variable
𝜕𝑟 θ
𝟏
𝐜𝐨𝐬 𝛉
𝒂𝒏𝒅
𝒙
=
𝐫𝟐
𝐫
𝑖. 𝑒 y =
sin θ
r
Diff partially w. r. t. r keep as 𝜃 constant
𝜕𝑥
( ) =−
sin θ
r2
𝜕𝑟 θ
𝜕𝑟
( )
𝜕𝑦 𝑥
r is function of 𝑥, 𝑦 variable
𝜕𝑟
1
𝜕𝑦 𝑥
2
𝜕𝑟
1
𝜕𝑦 𝑥
2
𝒅
… … … … … … . (𝐶 )
−3
( ) = − (x2 + y 2 ) 2
𝜕
𝜕𝑦
𝟏
𝟏
𝒅𝒙 𝒙
𝒙𝟐
( )=−
r = (x2 + y 2 )−1/2
𝒅
(x2 + y 2 )
𝒅𝒙
𝒙𝒏 = 𝒏𝒙𝒏−𝟏
𝑑𝑥
𝑑𝑥
−3
( ) = − (x2 + y 2 ) 2 (0 + 2y)
1
1
= (− ) ( 2)
2
=−
r
−3
2
(
2 sin θ
r
𝐱𝟐 + 𝐲𝟐 =
)
𝟏
𝐫𝟐
𝒂𝒏𝒅
𝒚=
𝐬𝐢𝐧 𝛉
𝐫
sin θ
r −2
𝜕𝑟
( ) = −r 2 sin θ … … … (D)
𝜕𝑦 𝑥
From (A), (B),(C) and (D)
𝜕𝑥
𝜕𝑟
𝜕𝑦
𝜕𝑟
( ) ( ) + ( ) ( ) = (−
𝜕𝑟 θ 𝜕𝑥 𝑦
𝜕𝑟 θ 𝜕𝑦 𝑥
𝜕𝑥
𝜕𝑦
𝜕𝑟
cos θ
r2
) (−r 2 cos θ) + (−
sin θ
r2
) (−r 2 sin θ)
𝜕𝑟
( ) ( ) + ( ) ( ) = cos 2 θ + sin2 θ = 1
𝜕𝑟 θ 𝜕𝑥 𝑦
𝜕𝑟 θ 𝜕𝑦 𝑥
**********
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Composite Function:
A composite function is a function that depends on another function.
The chain rule exists for differentiating a function of another function.
The chain rule is a formula to compute the derivative of a composite function. If
𝑓(𝑥, 𝑦) and 𝑥 = ℎ(𝑟, 𝜃), 𝑦 = g(𝑟, 𝜃) are differentiable functions.
𝑓(𝑥, 𝑦) is differentiable functions,
Diff partially w. r. t. to 𝑥
𝜕𝑓
𝜕𝑥
Diff partially w. r. t. to 𝑥, 𝑦
𝜕𝑓
𝜕𝑦
x, y are differentiable function,
Diff partially w. r. t. to 𝑟, 𝜃
𝜕𝑥
𝜕𝑟
,
𝜕𝑥
𝜕𝜃
and
𝜕𝑦
𝜕𝑟
,
𝜕𝑦
𝜕𝜃
By Chain
𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦
=
+
𝜕𝑟 𝜕𝑥 𝜕𝑟 𝜕𝑦 𝜕𝑟
𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦
=
+
𝜕𝜃 𝜕𝑥 𝜕𝜃 𝜕𝑦 𝜕𝜃
If 𝑓 = 𝑓1 (𝑥, 𝑦, 𝑧) 𝑎𝑛𝑑 𝑥, 𝑦, 𝑧 = 𝑓2 (𝑢, 𝑣, 𝑤 ) then
𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑧
=
+
+
𝜕𝑢 𝜕𝑥 𝜕𝑢 𝜕𝑦 𝜕𝑢 𝜕𝑧 𝜕𝑢
𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑧
=
+
+
𝜕𝑣 𝜕𝑥 𝜕𝑣 𝜕𝑦 𝜕𝑣 𝜕𝑧 𝜕𝑣
𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦 𝜕𝑓 𝜕𝑧
=
+
+
𝜕𝑤 𝜕𝑥 𝜕𝑤 𝜕𝑦 𝜕𝑤 𝜕𝑧 𝜕𝑤
****************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Examples 1: If z = 𝑓(u, v) and
u = x cos t – y sin t , v = x sin t + y cos t,
∂z
∂z
∂z
∂z
where t is a constant, prove that ∶ x + y = u + v
∂x
∂y
∂u
∂v
Solution: 𝑧 = 𝑓(𝑢, 𝑣)
𝑢 = 𝑥 𝑐𝑜𝑠 𝑡 – 𝑦 𝑠𝑖𝑛 𝑡 … … … (1)
𝑣 = 𝑥 𝑠𝑖𝑛 𝑡 + 𝑦 𝑐𝑜𝑠 𝑡 … … … (2)
u and v is differentiable functions of x and y
Diff u and v partially w. r. t. to x and y respectively
∂u
∂x
∂v
∂x
and
= 𝑠𝑖𝑛 𝑡
and
𝜕𝑧
By chain rule
𝜕𝑥
=
𝜕𝑧
𝜕𝑥
=
Multiplying by x
x
𝑥
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑦
𝜕𝑧
𝜕𝑦
𝜕𝑢 𝜕𝑥
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑥
𝜕𝑧 𝜕𝑢
𝜕𝑢 𝜕𝑦
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑦
+
𝜕𝑧
∂y
= cos 𝑡
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑢
sin 𝑡
𝜕𝑣
cos 𝑡 + 𝑥
+ 𝑥 sin 𝑡
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑣
sin 𝑡
… … … … … … … … … … (𝐴)
𝜕𝑧 𝜕𝑣
𝜕𝑣 𝜕𝑦
(− sin 𝑡) +
𝑦
= − 𝑠𝑖𝑛 𝑡
𝜕𝑣 𝜕𝑥
cos 𝑡 +
=𝑥
∂y
∂v
𝜕𝑧 𝜕𝑣
+
= 𝑥𝑐𝑜𝑠 𝑡
=
=
𝜕𝑧 𝜕𝑢
Multiplying by y
𝑦
∂u
= 𝑐𝑜𝑠 𝑡
𝜕𝑧
𝜕𝑦
=𝑦
= −𝑦 sin 𝑡
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑢
cos 𝑡
(− sin 𝑡) + 𝑦
+ 𝑦 cos 𝑡
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑣
cos 𝑡
… … … … … … … … … … … (𝐵)
Adding equations (A) and (B)
𝑥
𝑥
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑥
+ 𝑦
+ 𝑦
𝜕𝑧
𝜕𝑦
𝜕𝑧
𝜕𝑦
= 𝑥 𝑐𝑜𝑠 𝑡
𝜕𝑧
𝜕𝑢
+ 𝑥 sin 𝑡
= (𝑥𝑐𝑜𝑠 𝑡 − 𝑦 sin 𝑡 )
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑢
− 𝑦 sin 𝑡
𝜕𝑧
𝜕𝑢
+ 𝑦 cos 𝑡
+ (𝑥 sin 𝑡 + 𝑦 cos 𝑡 )
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑣
From (1) and (2)
𝑥
𝜕𝑧
𝜕𝑥
+ 𝑦
𝜕𝑧
𝜕𝑦
= 𝑢
𝜕𝑧
𝜕𝑢
+𝑣
𝜕𝑧
𝜕𝑣
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
*****************
Examples 2: 𝐼𝑓 𝑧 = 𝑓(𝑥, 𝑦) 𝑤ℎ𝑒𝑟𝑒 𝑥 = 𝑢 + 𝑣 𝑎𝑛𝑑 𝑦 = 𝑢𝑣
𝜕𝑧
𝜕𝑧
𝜕𝑧
𝜕𝑧
𝑡ℎ𝑒𝑛 𝑝𝑟𝑜𝑣𝑒 𝑡ℎ𝑎𝑡 ∶ 𝑢
+𝑣
=𝑥
+ 2𝑦
𝜕𝑢
𝜕𝑣
𝜕𝑥
𝜕𝑦
Solution: 𝑧 = 𝑓(𝑥, 𝑦) 𝑖. 𝑒. 𝑧 𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥, 𝑦
𝑥 = 𝑢+𝑣
… … … (1)
𝑦 = 𝑢𝑣
… … … . (2)
x and y is differentiable functions of u and v
Diff partially w. r. t. to u and v respectively
∂x
∂u
∂y
∂u
=1
𝑎𝑛𝑑
=𝑣
𝑎𝑛𝑑
𝜕𝑧
By chain rule
=
𝜕𝑢
𝜕𝑧
𝜕𝑢
=
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑣
=
=
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑧 𝜕𝑥
𝜕𝑥 𝜕𝑣
𝜕𝑧
𝜕𝑥
+
1 +
𝜕𝑧
𝑣
𝜕𝑣
=𝑣
𝜕𝑧
𝜕𝑥
1 + 𝑢
+ 𝑢𝑣
𝜕𝑥
𝜕𝑧
𝜕𝑥
=𝑢
𝑣
𝜕𝑦
=𝑢
𝜕𝑢
=𝑢
Multiplying by v
𝑣
𝜕𝑧
∂v
=1
𝜕𝑦 𝜕𝑢
𝜕𝑧
1 +
∂v
∂y
𝜕𝑧 𝜕𝑦
+
𝜕𝑥 𝜕𝑢
𝑢
Multiplying by u
𝑢
𝜕𝑧 𝜕𝑥
∂x
𝜕𝑧
𝜕𝑦
𝜕𝑧
𝑣
… … … … … … … … … … (𝐴)
𝜕𝑦
𝜕𝑧 𝜕𝑦
𝜕𝑦 𝜕𝑣
𝜕𝑧
𝜕𝑦
𝑢
=𝑣
𝜕𝑧
𝜕𝑥
+ 𝑢𝑣
1 +𝑣
𝜕𝑧
𝜕𝑧
𝜕𝑦
𝑢
… … … … … … … … … … … (𝐵)
𝜕𝑦
Adding equations (A) and (B)
𝑢
𝑢
𝜕𝑧
𝜕𝑢
𝜕𝑧
𝜕𝑢
+ 𝑣
+ 𝑣
𝜕𝑧
𝜕𝑣
𝜕𝑧
𝜕𝑣
=𝑢
𝜕𝑧
𝜕𝑥
+ 𝑢𝑣
= (𝑢 + 𝑣)
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑦
+ 𝑣
𝜕𝑧
𝜕𝑥
+ 𝑢𝑣
+ (𝑢𝑣 + 𝑢𝑣)
𝜕𝑧
𝜕𝑦
𝜕𝑧
𝜕𝑦
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
𝑢
𝜕𝑧
𝜕𝑢
+ 𝑣
Engineering Mathematics I
𝜕𝑧
𝜕𝑣
= (𝑢 + 𝑣)
𝜕𝑧
+ 2𝑢𝑣
𝜕𝑥
𝜕𝑧
𝜕𝑦
From (1) and (2)
𝑢
𝜕𝑧
𝜕𝑢
+ 𝑣
𝜕𝑧
𝜕𝑣
=𝑥
𝜕𝑧
𝜕𝑥
+ 2𝑦
𝜕𝑧
𝜕𝑦
***********
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Homogeneous Function:
A function 𝐹(𝑥, 𝑦) is said to be homogeneous function in which the power of
each term is same.
𝐹 (𝑥, 𝑦, 𝑧) = 𝑥 3 + 𝑦 3 + 𝑧 3 − 9𝑥 1 𝑦 2 + 19𝑥 2 𝑦1 − √3𝑦 2 𝑧1
Note:
o
To verify homogeneous function put 𝑥 = 𝑥𝑡, 𝑦 = 𝑦𝑡 𝑎𝑛𝑑 𝑧 = 𝑧𝑡
𝐹(𝑥, 𝑦, 𝑧) = (𝑥𝑡)3 + (𝑦𝑡)3 + (𝑧𝑡)3 − 9𝑥𝑡 1 𝑦𝑡 2 + 19𝑥𝑡 2 𝑦𝑡 1 − √3𝑦𝑡 2 𝑧𝑡 1
𝐹(𝑥, 𝑦, 𝑧) = 𝑥 3 𝑡 3 + 𝑦 3 𝑡 3 + 𝑡 3 𝑧 3 − 9𝑥𝑦 2 𝑡 3 + 19𝑥 2 𝑡 3 𝑦 − √3𝑦 2 𝑡 3 𝑧
𝐹 (𝑥, 𝑦, 𝑧) = 𝑡 3 𝐹(𝑥, 𝑦, 𝑧)
F is homogeneous function of degree 3
o
𝑓(𝑥, 𝑦, 𝑧) = 𝑡 𝑛 𝑓(𝑥, 𝑦, 𝑧) 𝑓 is homogeneous function of degree n
**********
Euler’s Theorem of Homogeneous Function:
Let 𝑓 (𝑥, 𝑦) be a homogenous function of degree n in x and y
For 2 variables and 1st order
𝑥
𝜕𝑓
𝜕𝑥
+𝑦
𝜕𝑓
𝜕𝑦
= 𝑛 𝑓(𝑥, 𝑦)
Let 𝑓 (𝑥, 𝑦, 𝑧) be a homogenous function of degree n so that 𝑓(𝑥, 𝑦, 𝑧) = 𝑡 𝑛 𝑓(𝑥, 𝑦, 𝑧)
For 3 variables and 1st order
𝑥
𝜕𝑓
𝜕𝑥
+𝑦
𝜕𝑓
𝜕𝑦
+𝑧
𝜕𝑓
𝜕𝑧
= 𝑛𝑓(𝑥, 𝑦, 𝑧)
Let 𝑓 be a homogenous function of degree n in 𝑥1 , 𝑥2 , 𝑥3 , … … . 𝑥𝑛
For n variables and 1st order
𝑥1
𝜕𝑓
𝜕𝑓
𝜕𝑓
𝜕𝑓
+ 𝑥2
+ 𝑥3
+ … … … … + 𝑥𝑛
=𝑛𝑓
𝜕𝑥1
𝜕𝑥2
𝜕𝑥3
𝜕𝑥𝑛
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Euler’s Theorem for second order
Let 𝑓(𝑥, 𝑦) be a homogenous function of two variables x, y of degree n
𝑥2
For 2 variables and 2nd order
𝜕2 𝑓
𝜕2 𝑓
𝜕𝑥
𝜕𝑥𝜕𝑦
2 + 2𝑥𝑦
+ 𝑦2
𝜕2 𝑓
𝜕𝑦 2
= 𝑛(𝑛 − 1)𝑓
******************
Example: If 𝒇(𝒙, 𝒚) =
𝒙
𝝏𝒇
𝝏𝒙
+𝒚
𝝏𝒇
𝝏𝒚
𝟏
𝒙𝟐
+
𝟏
𝐥𝐨𝐠 𝒙 − 𝐥𝐨𝐠 𝒚
+
𝒙𝒚
+ 𝟐𝒇 = 𝟎
Solution: Consider 𝑓(𝑥, 𝑦) =
𝑂𝑅
1
1
𝑥
𝑥𝑦
+
2
+
𝑥𝑦 + 𝑥 2
𝑓(𝑥, 𝑦) =
then prove that
𝒙𝟐 + 𝒚𝟐
𝑥3𝑦
𝑦+𝑥
𝑓(𝑥, 𝑦) =
𝑥2𝑦
𝐹𝑖𝑛𝑑 𝒙
log 𝑥 − log 𝑦
+
+
𝝏𝒇
𝝏𝒙
+𝒚
𝝏𝒇
𝝏𝒚
𝒂
∵ 𝐥𝐨𝐠 𝒂 − 𝐥𝐨𝐠 𝒃 = 𝐥𝐨𝐠 ( )
𝑥 2 +𝑦 2
𝒃
𝑥
𝑦
log( )
𝑥 2 +𝑦 2
𝑥
)
𝑦
𝑥 2 +𝑦 2
log (
𝑥
Let 𝑢 =
𝑦+𝑥
𝑥2𝑦
𝑣=
log(𝑦)
𝑥 2 +𝑦 2
𝑓(𝑥, 𝑦) = 𝑢 + 𝑣 … … … (𝐴)
Put
𝑥 = 𝑥𝑡, 𝑦 = 𝑦𝑡
𝑥𝑡
𝑢=
𝑦𝑡 + 𝑥𝑡
(𝑥𝑡)2 𝑦𝑡
𝑣=
log (𝑦𝑡)
(𝑥𝑡)2 + (𝑦𝑡)2
𝑣=
log (𝑦)
𝑡 2 (𝑥 2 + 𝑦 2 )
𝑥
𝑢=
𝑢=
𝑡(𝑦 + 𝑥)
𝑡3 𝑥2 𝑦
(𝑦 + 𝑥)
𝑡2 𝑥2 𝑦
𝑣=
𝑥
𝑦
log ( )
𝑡 2 (𝑥 2 + 𝑦 2 )
𝑥
𝑢=
(𝑦 + 𝑥)
𝑡 −2 2
𝑥 𝑦
𝑢 = 𝑡 −2 𝑢(𝑥, 𝑦)
𝑣 = 𝑡 −2
log (𝑦)
(𝑥 2 + 𝑦 2 )
𝑣 = 𝑡 −2 𝑣(𝑥, 𝑦)
∴ u and v are homogeneous function of degree n = −2
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
∴ by Eulers theorem
𝑥
𝜕𝑢
𝜕𝑢
+𝑦
= 𝑛𝑢
𝜕𝑥
𝜕𝑦
⇒ 𝑥
𝜕𝑢
𝜕𝑢
+𝑦
= −2𝑢
𝜕𝑥
𝜕𝑦
𝑥
𝜕𝑣
𝜕𝑣
+𝑦
= 𝑛𝑣
𝜕𝑥
𝜕𝑦
⇒ 𝑥
𝜕𝑣
𝜕𝑣
+𝑦
= −2𝑣
𝜕𝑥
𝜕𝑦
𝑓 =𝑢+𝑣
𝜕𝑓
𝜕𝑥
𝜕𝑢
=𝑥
𝜕𝑥
… … … (2)
𝑑𝑖𝑓𝑓 𝑤. 𝑟. 𝑡 𝑥 , 𝑦
𝜕𝑓 𝜕𝑢 𝜕𝑣
=
+
𝜕𝑥 𝜕𝑥 𝜕𝑥
𝑥
… … … (1)
𝜕𝑓 𝜕𝑢 𝜕𝑣
=
+
𝜕𝑦 𝜕𝑦 𝜕𝑦
&
𝜕𝑣
+𝑥
𝑦
𝜕𝑥
𝜕𝑓
=𝑦
𝜕𝑦
𝜕𝑢
+𝑦
𝜕𝑦
𝜕𝑣
𝜕𝑦
∴ Adding (1) and(2)
𝜕𝑢
⇒ 𝑥
⇒ 𝑥
⇒𝑥
𝜕𝑥
𝜕𝑢
𝜕𝑥
𝜕𝑓
𝜕𝑥
⇒ 𝑥
+𝑥
+𝑦
𝜕𝑓
𝜕𝑥
𝜕𝑢
+𝑦
𝜕𝑦
𝜕𝑣
𝜕𝑥
𝜕𝑓
+𝑦
𝜕𝑣
𝜕𝑥
𝜕𝑢
𝜕𝑦
+𝑦
+𝑦
𝜕𝑣
= −2𝑢 − 2𝑣
𝜕𝑦
𝜕𝑣
= −2(𝑢 + 𝑣)
𝜕𝑦
= −2𝑓
𝜕𝑦
+𝑦
+𝑥
𝜕𝑓
𝜕𝑦
+ 2𝑓 = 0
**********
Example: If 𝑢 =
𝑥
𝜕𝑓
𝜕𝑥
+𝑦
𝑥3 + 𝑦3
𝑥+𝑦
𝜕𝑓
𝜕𝑦
1
+
+ x2
𝑥5
𝑥2 + 𝑦2
𝑥 2 + 2𝑥𝑦
∂2 f
∂2 f
∂x
∂x ∂y
+ 2xy
2
Solution: Consider 𝑓 (𝑥, 𝑦) =
Let
sin−1 (
𝑥3 + 𝑦3
𝑥+𝑦
𝑢=
+
𝑥3 + 𝑦3
𝑥+𝑦
) then find the value of
+ y2
1
𝑥5
∂2 f
at the point (1, 2)
∂y2
−1
sin
(
𝑣=
𝑥2 + 𝑦2
𝑥 2 + 2𝑥𝑦
1
𝑥5
)
sin−1 (
𝑥2 + 𝑦2
𝑥 2 + 2𝑥𝑦
)
𝑓 (𝑥, 𝑦) = 𝑢 + 𝑣 … … … (𝐴)
Put
𝑢=
𝑥 = 𝑥𝑡, 𝑦 = 𝑦𝑡
(𝑥𝑡)3 + (𝑦𝑡)3
𝑥𝑡+ 𝑦𝑡
𝑣=
1
(𝑥𝑡)5
sin−1 (
(𝑥𝑡)2 + (𝑦𝑡)2
(𝑥𝑡)2 + 2𝑥𝑡𝑦𝑡
)
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
𝑡 3 (𝑥 3 + 𝑦 3 )
𝑢=
𝑡(𝑥 + 𝑦)
3
3
2 (𝑥 + 𝑦 )
𝑡
(𝑥 + 𝑦)
𝑢=
1
𝑣=
𝑥5 𝑡 5
𝑣=
𝑢 = 𝑡 2 𝑢(𝑥, 𝑦)
sin−1 (
1
𝑡 −5 5
𝑥
𝑡 2 (𝑥 2 + 𝑦 2 )
𝑡 2 (𝑥 2 + 2𝑥𝑦)
−1
sin
(
𝑥2 + 𝑦2
𝑥 2 + 2𝑥𝑦
)
)
𝑣 = 𝑡 −5 𝑣(𝑥, 𝑦)
∴ u is homogeneous function of degree n = 2
∴ v is homogeneous function of degree n = −5
∴ by Eulers theorem
𝜕𝑢
𝑥
𝜕𝑥
⇒ 𝑥
x2
𝜕𝑢
𝜕𝑥
+𝑦
𝜕𝑓
𝜕𝑦
=
𝜕𝑥
= 2𝑢
∂x
∂x ∂y
2
∂2 u
+ 2xy
𝜕𝑥
+𝑦
𝜕𝑣
𝜕𝑦
∂x ∂y
+ y2
+
… … … (1)
∂2 u
= n(n − 1)u
∂y2
2
∂ u
y2 2
∂y
∂2 v
∂x
∂x ∂y
2
∂2 v
+
𝜕𝑢
𝜕𝑦
= −5𝑣
∂2 v
+ 2xy
2
= 2(2 − 1)𝑢
= 2𝑢
… … … . (2)
𝜕𝑣
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+ y2
… … … (3)
∂2 v
∂y2
2
∂ v
2xy
+ y2 2
∂x ∂y
∂y
𝑓 = 𝑢+𝑣
𝜕𝑥
𝜕𝑦
∂2 u
∂ v
x2 2 +
∂x
𝜕𝑢
𝜕𝑢
2 + 2xy
𝜕𝑣
𝑥
x2
=
= 𝑛𝑢
𝜕𝑣
𝜕𝑣
+𝑦
= 𝑛𝑣
𝜕𝑥
𝜕𝑦
𝑥
𝜕𝑓
𝜕𝑦
∂2 u
∂ u
x2 2
∂x
⇒
𝜕𝑢
+𝑦
= n(n − 1)𝑣
= −5(−5 − 1)𝑣 = 30𝑣
… … … (4)
𝑑𝑖𝑓𝑓 𝑤. 𝑟. 𝑡 𝑥 , 𝑦
multiply by 𝑥
𝑥
multiply by 𝑦
𝑦
𝜕𝑓
𝜕𝑥
=𝑥
𝜕𝑓
𝜕𝑦
𝜕𝑢
𝜕𝑥
=𝑦
+𝑥
𝜕𝑢
𝜕𝑦
𝜕𝑣
𝜕𝑥
+𝑦
𝜕𝑣
𝜕𝑦
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
𝜕2 𝑓
𝜕𝑥 2
=
𝜕2 𝑓
𝜕𝑦 2
=
𝜕2 𝑓
𝜕𝑥𝜕𝑦
Engineering Mathematics I
𝜕2 𝑢
𝜕2 𝑣
𝜕𝑥
𝜕𝑥 2
2 +
𝜕2 𝑢
𝜕𝑦 2
=
+
𝜕2 𝑢
𝜕𝑥𝜕𝑦
multiply by 𝑥 2
𝜕2 𝑣
multiply by 𝑦
𝜕𝑦 2
𝜕2 𝑣
+
𝜕𝑥𝜕𝑦
𝜕2 𝑓
𝑥2
𝜕𝑥 2
2
2 𝜕 𝑓
𝑦
𝜕𝑦 2
2
multiply by 2𝑥𝑦
= 𝑥2
𝜕2 𝑣
𝜕𝑥
𝜕𝑥 2
2
2 + 𝑥
2
𝜕 𝑢
𝑦2 2
𝜕𝑦
=
𝜕2 𝑓
2𝑥𝑦
𝜕2 𝑢
= 2𝑥𝑦
𝜕𝑥𝜕𝑦
+
2
2𝜕 𝑣
𝑦
𝜕𝑦 2
𝜕2 𝑢
𝜕𝑥𝜕𝑦
+ 2𝑥𝑦
𝜕2 𝑣
𝜕𝑥𝜕𝑦
∴ Adding (1), (2), (3)and (4)
𝜕𝑢
𝑥
𝜕𝑥
+𝑦
𝜕𝑢
𝜕𝑦
+ +x 2
∂2 u
∂2 u
∂x
∂x ∂y
+ 2xy
2
+ y2
∂2 u
𝜕𝑣
∂y
𝜕𝑥
+𝑥
2
+𝑦
𝜕𝑣
𝜕𝑦
+ x2
∂2 v
∂2 v
∂x
∂x ∂y
+ 2xy
2
+ y2
∂2 v
∂y2
= 2𝑢 − 5𝑣 + 2𝑢 + 30𝑣
𝑥
𝑥
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑥
+𝑦
+𝑦
𝜕𝑓
𝜕𝑦
𝜕𝑓
𝜕𝑦
+
2
2∂ f
x 2
∂x
+
2
2∂ f
x 2
∂x
∂2 f
+ 2xy
∂x ∂y
∂2 f
+ 2xy
∂x ∂y
+
2
2∂ f
y
∂y2
+
2
2∂ f
y
∂y2
= 4𝑢 + 25𝑣
𝑥3 + 𝑦3
=4
𝑥+𝑦
+ 25
1
−1
𝑥5
sin
(
𝑥2 + 𝑦2
𝑥 2 + 2𝑥𝑦
)
At point (1, 2) put x = 1 and y = 2
𝑥
𝜕𝑓
𝜕𝑥
𝑥
𝑥
+
𝜕𝑓
𝜕𝑥
𝜕𝑓
𝜕𝑥
2
2∂ f
𝑦 +x 2
𝜕𝑦
∂x
𝜕𝑓
+𝑦
+𝑦
𝜕𝑓
𝜕𝑦
𝜕𝑓
𝜕𝑦
+ x2
+ x2
+
∂2 f
2
2 ∂ f
2xy
+y
∂x ∂y
∂y2
∂2 f
∂2 f
∂x
∂x ∂y
∂2 f
∂2 f
+ 2xy
2
2 + 2xy
∂x
∂x ∂y
+ y2
+ y2
=4
∂2 f
∂y2
∂2 f
∂y2
13 + 23
1+2
+ 25
1
15
−1
sin
9
5
3
5
(
12 + 22
12 + 2(1)(2)
= 4 ( ) + 25 sin−1 ( )
= 12 + 25
𝜋
2
**************
𝑦
𝑥
𝑥
𝑦
Example: If 𝑢 = 𝑥 𝑛 𝑓 ( ) + 𝑦 −𝑛 g ( ) then
𝑥
𝜕𝑓
𝜕𝑥
+𝑦
𝜕𝑓
𝜕𝑦
+ x2
∂2 f
2 + 2xy
∂x
∂2 f
∂x ∂y
+ y2
∂2 f
∂y2
= n2 u
**************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
)
Prof. N. R. Kharat
Engineering Mathematics I
Euler’s Theorem of Non-Homogeneous Function:
Let 𝑢 = 𝑓 −1 (𝑥, 𝑦) be a non-homogenous function but 𝑓(𝑢) is homogenous
function of order n so that 𝑓(𝑢) = 𝑡 𝑛 𝑓(𝑢)
𝜕𝑓
For 2 variables and 1st order
𝑥
For 2 variables and 2nd order
𝑥2
𝜕𝑥
+𝑦
𝜕𝑓
=𝑛
𝜕𝑦
𝑓(𝑢)
𝑓′ (𝑢)
𝜕2 𝑓
𝜕2 𝑓
𝜕𝑥
𝜕𝑥𝜕𝑦
+ 2𝑥𝑦
2
= g(𝑢)
+ 𝑦2
Where g(𝑢) = 𝑛
𝜕2 𝑓
𝜕𝑦 2
= g(𝑢 )[g′ (𝑢) − 1]
𝑓(𝑢)
𝑓′ (𝑢)
*************
Example: If 𝑢 = sin−1(𝑥 2 + 𝑦 2 )1/5 then Prove that
∂2 f
∂2 f
∂2 f 2 tan u 2 tan2 u 3
2
[
x
+ 2xy
+y
=
− ]
∂x2
∂x ∂y
∂y 2
5
5
5
2
Solution: Consider 𝑢(𝑥, 𝑦) = sin−1(𝑥 2 + 𝑦 2 )1/5
𝑥 = 𝑥𝑡, 𝑦 = 𝑦𝑡
Put
𝑢 = sin−1[(𝑥𝑡)2 + (𝑦𝑡)2 ]1/5
𝑢 = sin−1[𝑡 2 (𝑥 2 + 𝑦 2 )]1/5
(𝒂𝒏 )𝒎 = 𝒂𝒏𝒎
𝑢 = sin−1[𝑡 2/5 (𝑥 2 + 𝑦 2 )1/5 ]
𝐬𝐢𝐧−𝟏(𝒂𝒙) ≠ 𝒂 𝐬𝐢𝐧−𝟏(𝒙)
𝑢(𝑥, 𝑦) ≠ 𝑡 𝑛 𝑢(𝑥, 𝑦) ∴ u is not homogeneous function
sin 𝑢 = 𝑡 2/5 (𝑥 2 + 𝑦 2 )1/5
∴ 𝑓(u) = sin u is homogeneous function of degree n =
𝑥2
𝜕2 𝑓
𝜕𝑥
2 + 2𝑥𝑦
𝜕2 𝑓
+ 𝑦2
𝜕𝑥𝜕𝑦
where g(𝑢) = 𝑛
𝜕𝑦 2
= g(u)[g ′ (u) − 1]
5
… … … . (𝐴)
𝑓(𝑢)
𝑓′ (𝑢)
here 𝑓(𝑢) = 𝑠𝑖𝑛 𝑢
⇒ g(𝑢) = 𝑛
𝜕2 𝑓
2
𝑓(𝑢)
𝑓′ (𝑢)
=
⇒ 𝑓 ′ (𝑢) = 𝑐𝑜𝑠 𝑢
2 𝑠𝑖𝑛 𝑢
5 𝑐𝑜𝑠 𝑢
=
2
5
tan 𝑢
2
g′ (𝑢) = 𝑠𝑒𝑐 2 𝑢
5
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
Equation (A)
𝑥2
𝜕2 𝑓
𝜕2 𝑓
𝜕𝑥
𝜕𝑥𝜕𝑦
+ 2𝑥𝑦
2
2
𝜕 𝑓
𝑥2 2
𝜕𝑥
2
𝜕 𝑓
𝑥2 2
𝜕𝑥
𝑥2
𝑥2
+ 2𝑥𝑦
+ 2𝑥𝑦
𝜕2 𝑓
𝜕𝑥𝜕𝑦
𝜕2 𝑓
𝜕𝑥𝜕𝑦
𝜕2 𝑓
𝜕2 𝑓
𝜕𝑥
𝜕𝑥𝜕𝑦
+ 2𝑥𝑦
2
𝜕2 𝑓
𝜕𝑥
+ 2𝑥𝑦
2
𝜕2 𝑓
𝜕𝑥𝜕𝑦
+ 𝑦2
𝜕2 𝑓
𝜕𝑦 2
+
2
2𝜕 𝑓
𝑦
𝜕𝑦 2
+
2
2𝜕 𝑓
𝑦
𝜕𝑦 2
+ 𝑦2
+ 𝑦2
2
2𝑠𝑒𝑐 2 𝑢
5
5
= tan 𝑢 [
2
2(𝑡𝑎𝑛 2 𝑢 + 1)
5
5
2
2 𝑡𝑎𝑛 2 𝑢 + 2
5
5
= tan 𝑢 [
= tan 𝑢 [
𝜕2 𝑓
𝜕2 𝑓
− 1]
2 𝑡𝑎𝑛 2 𝑢 + 2 − 5
5
5
2
2 𝑡𝑎𝑛 2 𝑢
5
5
= tan 𝑢 [
𝜕𝑦 2
− 1]
2
= tan 𝑢 [
𝜕𝑦 2
− 1]
]
3
− ]
5
********
𝑥 1/2 + 𝑦 1/2
Example: If 𝑢 = cosec −1 √
𝑥 1/3 + 𝑦 1/3
then Prove that
∂2 u
∂2 u
∂2 u tan u tan2 u 13
2
[
x
+ 2xy
+y
=
+ ]
∂x2
∂x ∂y
∂y 2
12
12
12
2
Solution: Consider 𝑢 =
Put
1/2 + 𝑦 1/2
−1 √𝑥
cosec
𝑥 1/3 + 𝑦 1/3
𝑥 = 𝑥𝑡, 𝑦 = 𝑦𝑡
(𝑥𝑡)1/2 + (𝑦𝑡)1/2
(𝑥𝑡)1/3 + (𝑦𝑡)1/3
𝑢 = cosec
−1 √
𝑢 = cosec
−1 √
𝑡 1/2 (𝑥 1/2 + 𝑦1/2 )
𝑡 1/3 (𝑥 1/3 + 𝑦1/3 )
1/2
𝑡 1/2 (𝑥 1/2 + 𝑦1/2 )
−1
𝑢 = cosec [ 1/3 1/3
]
𝑡 (𝑥
+ 𝑦1/3 )
1/2
𝑡 1/2 𝑡 −1/3 (𝑥1/2 + 𝑦1/2 )
−1
𝑢 = cosec [
]
(𝑥 1/3 + 𝑦1/3 )
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
Other Subjects: https://www.studymedia.in/fe/notes
Prof. N. R. Kharat
Engineering Mathematics I
1/2
𝑡 1/6 (𝑥 1/2 + 𝑦1/2 )
−1
𝑢 = cosec [
]
(𝑥 1/3 + 𝑦1/3 )
𝑢 = cosec
−1
[𝑡
𝑥1/2 + 𝑦1/2
]
𝑥1/3 + 𝑦1/3
1/12 √
𝑢(𝑥, 𝑦) ≠ 𝑡 𝑛 𝑢(𝑥, 𝑦) ∴ u is not homogeneous function
𝑥1/2 + 𝑦1/2
𝑥1/3 + 𝑦1/3
cosec 𝑢 = 𝑡1/12 √
∴ 𝑓(u) = cosec u is homogeneous function of degree n =
2
2∂ u
x 2+
∂x
2xy
∂2 u
∂x ∂y
2
∂ u
y2 2
∂y
+
= g(u)[g ′ (u) − 1]
𝑤ℎ𝑒𝑟𝑒 g(𝑢) = 𝑛
x2
x2
12
∂2 u
∂x
∂x ∂y
∂2 u
∂2 u
∂x
∂x ∂y
2 + 2xy
2
2∂ u
x 2
∂x
+ y2
+ y2
∂2 u
∂2 u
∂y2
∂2 u
∂y2
2
∂2 u
2
2∂ u
2xy
+y 2
∂x ∂y
∂y
∂2 u
∂2 u
∂x
∂x ∂y
2 + 2xy
1
𝑐𝑜𝑠𝑒𝑐 𝑢
12 𝑐𝑜𝑠𝑒𝑐 𝑢 cot 𝑢
=−
1
tan 𝑢
12
𝑠𝑒𝑐 2 𝑢
∂ u
+ 2xy
+ y2 2
∂x ∂y
∂y
2
2∂ u
x 2+
∂x
x2
1
∂2 u
2 + 2xy
= −
𝑓′ (𝑢)
1 1
12 cot 𝑢
g(𝑢) = −
g′ (𝑢) = −
𝑓(𝑢)
𝑓′ (𝑢)
⇒ 𝑓 ′ (𝑢) = −𝑐𝑜𝑠𝑒𝑐 𝑢 cot 𝑢
here 𝑓(𝑢) = 𝑐𝑜𝑠𝑒𝑐 𝑢
⇒ g(𝑢) = 𝑛
𝑓(𝑢)
+ y2
∂2 u
∂y2
=−
=
=
=
=
1
12
1
12
1
12
1
12
1
12
tan 𝑢 [
tan 𝑢 [
tan 𝑢 [
tan 𝑢 [
tan 𝑢 [
−𝑠𝑒𝑐 2 𝑢
12
𝑠𝑒𝑐 2 𝑢
12
− 1]
+ 1]
𝑡𝑎𝑛 2 𝑢 + 1
12
+ 1]
𝑡𝑎𝑛 2 𝑢 + 1 + 12
]
12
𝑡𝑎𝑛 2 𝑢 + 13
12
]
************
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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1
12
Prof. N. R. Kharat
Engineering Mathematics I
𝑥+𝑦
Example: If 𝑢 = sec −1 [
𝑥 1/2
+ 𝑦 1/2
] then Prove that
∂2 u
∂2 u
∂2 u
cot u
2
[3 + cot2 u]
x
+
2xy
+
y
=
−
2
2
∂x
∂x ∂y
∂y
4
************
2
𝑥3 + 𝑦3
Example: If 𝑢 = tan−1 [
𝑥+ 𝑦
] then Prove that
∂2 u
∂2 u
∂2 u
2
x
+ 2xy
+y
= sin 2u [1 − 4 sin2 𝑢]
2
2
∂x
∂x ∂y
∂y
************
2
𝑥+𝑦
Example: If 𝑢 = sin−1 [
𝑥 1/2 + 𝑦 1/2
] then Prove that
∂2 f
∂2 f
∂2 f
−sin u cos 2u
2
x
+
2xy
+
y
=
∂x2
∂x ∂y
∂y 2
4 cos 3 u
************
2
𝑥2
Example: If 𝑢 = sin−1 [
1
2
+𝑦 2
𝑥+𝑦
]
then Prove that
∂2 u
∂2 u
∂2 u
tan u
2
[tan2 u − 1]
x
+
2xy
+
y
=
2
2
∂x
∂x ∂y
∂y
4
************
2
"𝑇ℎ𝑒 𝑂𝑛𝑙𝑦 𝑡ℎ𝑖𝑛𝑔𝑠 𝑡ℎ𝑎𝑡 𝑤𝑖𝑙𝑙 𝑠𝑡𝑜𝑝 𝑦𝑜𝑢 𝑓𝑟𝑜𝑚 𝑓𝑢𝑙𝑓𝑖𝑙𝑙𝑖𝑛𝑔 𝑦𝑜𝑢𝑟 𝑑𝑟𝑒𝑎𝑚𝑠 𝑖𝑠 𝑦𝑜𝑢"
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