3. EXACT DIFFERENTIAL EQUATION
A differential equation of the form
M(x, y)dx + N(x, y)dy = 0
is called an exact differential equation if and only
if
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4. SOLUTION OF EXACT D.E.
• The solution is given by :
𝑦=𝑐𝑜𝑛𝑠𝑡 𝑎 𝑛𝑡
𝑀𝑑𝑥 + 𝑡𝑒𝑟𝑚𝑠𝑜𝑓𝑁𝑛𝑜𝑡𝑐𝑜𝑛𝑡 𝑎𝑖 𝑛𝑖 𝑛 𝑔𝑥 𝑑𝑦 = 𝑐
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6. Example : 1 (cont.)
The given differential equation is exact ,
𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀𝑑𝑥 + 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑁 𝑛𝑜𝑡 𝑐𝑜𝑛𝑡𝑎𝑖 𝑛𝑖 𝑛 𝑔 𝑥 𝑑𝑦 = 𝑐
𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑦𝑒 𝑥 𝑑𝑥 + 2𝑦 𝑑𝑦 = 0
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⇒
⇒
⇒ 𝒚𝒆 𝒙+ 𝒚 𝟐= c
7.
8. NON EXACT DIFFERENTIAL
EQUATION
• For the differential equation
𝑀 𝑥, 𝑦 𝑑𝑥 + 𝑁 𝑥, 𝑦 𝑑𝑦 = 0
IF
𝝏𝑴
𝝏𝒚
≠
𝝏𝑵
𝝏𝒙
then,
𝑫𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕𝒊𝒂𝒍 𝑬𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝒊𝒔 𝒔𝒂𝒊𝒅 𝒕𝒐 𝒃𝒆 𝑵𝑶𝑵𝑬𝑿𝑨𝑪𝑻
• If the given differential equation is not exact then make
that equation exact by finding INTEGRATING
FACTOR.
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9. INTEGRATING FACTOR
• In general, for differential equation
M(x, y)dx + N(x, y)dy = 0
is not exact.
In such situation, we find a function 𝜆 such that by
multiplying𝜆 to the equation, it becomes an exact equation.
So,
𝝀M(x, y)dx +𝝀N(x, y)dy = 0 becomes exact equation
Here the function 𝜆 = 𝜆(𝑥, 𝑦) is then called an Integrating
Factor
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10. Methods to find an INTEGRATING
FACTOR (I.F.) for given non exact
equation:
M(x, y)dx + N(x, y)dy = 0
CASES:
CASE I
CASE II
CASE III
CASE IV
11. CASE I :
If
1
𝑁
(
𝜕𝑀
𝜕𝑦
+
𝜕𝑁
𝜕𝑥
)f 𝑥 {i.e. function of x only}
Then I.F. = 𝒆 𝒇 𝒙 .𝒅𝒙
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17. CASE II :
If
1
𝑀
(
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
) is a function of y only ,
say g(y),
then 𝑒 𝑔 𝑦 𝑑𝑦
is an I.F.(Integrating
Factor).
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18. Example : 3
Solve 𝑦4 + 2y dx + 𝑥𝑦3 + 2𝑦4 − 4𝑥 𝑑𝑦=0
Solution:
Here M=𝑦4 + 2𝑦 and so
𝜕𝑀
𝜕𝑦
= 4𝑦3+2
N=𝑥𝑦3 + 2𝑦4 − 4𝑥 and so
𝜕𝑁
𝜕𝑥
= 𝑦3 − 4
Thus,
𝜕𝑀
𝜕𝑌
≠
𝜕𝑁
𝜕𝑥
and so the given differential equation
is non exact.
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19. Example : 3 (cont.)
Now,
1
𝑀
𝜕𝑁
𝜕𝑥
−
𝜕𝑀
𝜕𝑦
=
1
𝑦4+2𝑦
𝑦3
− 4 − 4𝑦3
− 2
=
−(3𝑦3+6)
𝑦(𝑦3+2)
= -
3
𝑦
,
which is a function of y only . Therefore
I.F.=𝑒
−3
𝑦=𝑒−3𝑙𝑛𝑦 =
1
𝑦3
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20. Example : 3 (cont.)
Multiplying the given differential equation by
1
𝑦3 ,we have
1
𝑦3
𝑦4
+ 2𝑦 𝑑𝑥 + 𝑥𝑦3
+ 2𝑦4
− 4𝑥 𝑑𝑦
⇒ 𝑦 +
2
𝑦2 𝑑𝑥 + 𝑥 + 2𝑦 − 4
𝑥
𝑦3 𝑑𝑦 = 0 ----------------(i)
Now here, M=𝑦 +
2
𝑦2 and so
𝜕𝑀
𝜕𝑥
= 1 −
4
𝑦3
N=𝑥 + 2𝑦 − 4
𝑥
𝑦3 and so
𝜕𝑁
𝜕𝑥
= 1 −
4
𝑦3
Thus,
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
and hence 𝑒𝑞 𝑛
(i) is an exact
differential equatio
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21. Example : 3 (cont.)
Therefore , General Solution is
𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀𝑑𝑥 +
(𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒
𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥)
𝑑𝑦 = 𝑐
⇒ 𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑦 +
2
𝑦2 𝑑𝑥 + 2𝑦𝑑𝑦 = 𝑐
⇒𝒙𝒚 +
𝟐𝒙
𝒚 𝟐 + 𝒚 𝟐
=c
where c is an arbitrary constant.
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22. CASE III :
If the given differential equation is
homogeneous with 𝑀𝑥 + 𝑁𝑦 ≠
0 then
1
𝑀𝑥+𝑁𝑦
is an I.F.
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23. Example : 4
Solve 𝑥2
𝑦𝑑𝑥 − 𝑥3
+ 𝑥𝑦3
𝑑𝑦 = 0
Solution:
Here M=𝑥2 𝑦 and so
𝜕𝑀
𝜕𝑦
= 𝑥2
N=−𝑥3 − 𝑥𝑦2 and so
𝜕𝑁
𝜕𝑥
= −3𝑥2 − 𝑦2
∴
𝜕𝑀
𝜕𝑦
≠
𝜕𝑁
𝜕𝑥
∴ The given differential equation is non exact.
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24. Example : 4 (cont.)
The given differential equation is homogeneous
function of same degree=3.
[ 𝑀 𝑡𝑥, 𝑡𝑦 = (𝑡𝑥)2(𝑡𝑦)
=𝑡3 𝑥2 𝑦
=𝑡3
𝑀 𝑥, 𝑦
𝑁 𝑥, 𝑦 = −3 𝑡𝑥 3 − (𝑡𝑥) (𝑡𝑦)2
=𝑡3 −3𝑥2 − 𝑥𝑦2
=𝑡3 𝑁 𝑥, 𝑦
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26. Example : 4 (cont.)
⇒ −
𝑥
𝑦2
𝑑𝑥 +
𝑥3
𝑦3
+
1
𝑦
𝑑𝑦 = 0 −−−−−− −(i)
Here, M=
−𝑥
𝑦2 and so
𝜕𝑀
𝜕𝑦
=
2𝑥
𝑦3
N=
𝑥2
𝑦3 +
1
𝑦
and so
𝜕𝑁
𝜕𝑥
=
2𝑥
𝑦3
Thus,
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
and hence 𝑒𝑞 𝑛
(i) is an exact
differential equation.
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27. Example : 4 (cont.)
Therefore , General Solution is
𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀𝑑𝑥 +
(𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒
𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥)
𝑑𝑦 = 𝑐
⇒
𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
−𝑥
𝑦2
𝑑𝑥 +
1
𝑦
𝑑𝑦 = 𝑐
⇒ −
𝟏
𝟐
𝒙 𝟐
𝒚 𝟐
+ 𝒍𝒏𝒚 = 𝒄
where c is an arbitrary constant.
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28. CASE IV :
If the given differential equation is of the
form 𝑓1 𝑥𝑦 𝑦𝑑𝑥 + 𝑓2 𝑥𝑦 𝑥𝑑𝑦 =
0 𝑤𝑖𝑡ℎ 𝑀𝑥 − 𝑁𝑦 ≠ 0, 𝑡ℎ𝑒𝑛
1
𝑀𝑥−𝑁𝑦
is an
I.F.
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29. Example : 4 (cont.)
Solve (𝑥2 𝑦 + 2)𝑦𝑑𝑥 + 2 − 𝑥2 𝑦2 𝑥𝑑𝑦 = 0
Solution:
Here, M=(𝑥2 𝑦3 + 2𝑦) and so
𝜕𝑀
𝜕𝑦
= 3𝑥2 𝑦2 + 2
N=(𝑥 − 𝑥3 𝑦2) and so
𝜕𝑁
𝜕𝑥
= 2 − 3𝑥𝑦2
∴
𝜕𝑀
𝜕𝑦
≠
𝜕𝑁
𝜕𝑥
∴ The given differential equation is non exact.
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30. Example : 4 (cont.)
Now,
𝑀𝑥 − 𝑁𝑦 = 𝑥2
𝑦2
+ 2 𝑦𝑥 + 2 − 𝑥2
𝑦2
𝑥𝑦
=𝑥3 𝑦3 + 2𝑥𝑦 − 2𝑥𝑦 + 𝑥3 𝑦3
=2𝑥3
𝑦3
So, I.F.=
1
𝑀𝑥−𝑁𝑦
=
1
2𝑥3 𝑦3
Multiplying the given equation by
1
2𝑥3 𝑦3 , we have
1
2𝑥3 𝑦3
[ 𝑥2 + 2 𝑦𝑑𝑥 + 2 − 𝑥2 𝑦2 𝑥𝑑𝑦]
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31. Example : 4 (cont.)
⇒
1
2𝑥
+
1
𝑥3 𝑦3 𝑑𝑥 +
1
𝑥2 𝑦3 −
1
2𝑦
𝑑𝑦 = 0----------(i)
Here, M=
1
2𝑥
+
1
𝑥3 𝑦3 and so
𝜕𝑀
𝜕𝑦
= −
2
𝑥3 𝑦3
N=
1
𝑥2 𝑦3 −
1
2𝑦
and so
𝜕𝑁
𝜕𝑥
= −
2
𝑥3 𝑦3
Thus,
𝜕𝑀
𝜕𝑦
=
𝜕𝑁
𝜕𝑥
and hence 𝑒𝑞 𝑛
(i) is an exact
differential equation.
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32. Example : 4 (cont.)
Therefore , General Solution is
𝑦𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑀𝑑𝑥 +
(𝑇𝑒𝑟𝑚𝑠 𝑖𝑛 𝑁 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒
𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑜𝑓 𝑥)
𝑑𝑦 = 𝑐
⇒
𝑦 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
1
2𝑥
+
1
𝑥3 𝑦3
𝑑𝑥 + −
1
2𝑦
𝑑𝑦 = 𝑐
⇒
𝟏
𝟐
𝒍𝒐𝒈𝒙 −
𝟏
𝟐𝒙 𝟐 𝒚 𝟐
−
𝟏
𝟐
𝒍𝒐𝒈𝒚 = 𝒄
where c is an arbitrary constant.
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