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HOMOGENEOUS DE
Presented by :
MD.Hasnat Shoheb
ID : 152-15-5813
Daffodil International University
Department of CSE
HOMOGENEOUS DE
Question :
𝒅𝒚
𝒅𝒙
=
𝒚
𝒙+ 𝒙𝒚
SOLUTION : let us consider, y = vx ,
𝑑𝑦
𝑑𝑥
= 𝑣 + 𝑥
𝑑𝑣
𝑑𝑥
𝑣𝑥
𝑥+ 𝑥𝑣𝑥
= v + x
𝑑𝑣
𝑑𝑥
𝑣𝑥
𝑥+ 𝑥2 𝑣
= v+𝑥
𝑑𝑣
𝑑𝑥
SOLUTION :
𝑣𝑥
𝑥(1+ 𝑣)
= v + 𝑥
𝑑𝑣
𝑑𝑥
𝑣
1+√𝑣
− 𝑣 = 𝑥
𝑑𝑣
𝑑𝑥
−𝑣.𝑣
1
2
1+ √𝑣
= x
𝑑𝑣
𝑑𝑥
1
𝑥
𝑑𝑥 = −
1+ 𝑣 𝑑𝑣
𝑣
3
2
SOLUTION :
lnx = -
1
𝑣
3
2
dv -
𝑣
1
2
𝑣
3
2
dv
lnx = - 𝑣
−3
2 dv - 𝑣
1−3
2 dv
lnx = -
𝑣
−3+2
2
−3+2
2
- ln v
lnx = -
𝑣
−1
2
−1
2
- ln v
lnx = 2𝑣
−1
2 - ln v
SOLUTION :
ln x + ln v = 2𝑣
−1
2
ln (x+v) = 2 (
𝑦
𝑥
)
−1
2
ln [x+
𝑦
𝑥
] =2 (
𝑦
𝑥
)
−1
2
(Ans)
Mathmatics presentation1

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Mathmatics presentation1

  • 1. HOMOGENEOUS DE Presented by : MD.Hasnat Shoheb ID : 152-15-5813 Daffodil International University Department of CSE
  • 2. HOMOGENEOUS DE Question : 𝒅𝒚 𝒅𝒙 = 𝒚 𝒙+ 𝒙𝒚 SOLUTION : let us consider, y = vx , 𝑑𝑦 𝑑𝑥 = 𝑣 + 𝑥 𝑑𝑣 𝑑𝑥 𝑣𝑥 𝑥+ 𝑥𝑣𝑥 = v + x 𝑑𝑣 𝑑𝑥 𝑣𝑥 𝑥+ 𝑥2 𝑣 = v+𝑥 𝑑𝑣 𝑑𝑥
  • 3. SOLUTION : 𝑣𝑥 𝑥(1+ 𝑣) = v + 𝑥 𝑑𝑣 𝑑𝑥 𝑣 1+√𝑣 − 𝑣 = 𝑥 𝑑𝑣 𝑑𝑥 −𝑣.𝑣 1 2 1+ √𝑣 = x 𝑑𝑣 𝑑𝑥 1 𝑥 𝑑𝑥 = − 1+ 𝑣 𝑑𝑣 𝑣 3 2
  • 4. SOLUTION : lnx = - 1 𝑣 3 2 dv - 𝑣 1 2 𝑣 3 2 dv lnx = - 𝑣 −3 2 dv - 𝑣 1−3 2 dv lnx = - 𝑣 −3+2 2 −3+2 2 - ln v lnx = - 𝑣 −1 2 −1 2 - ln v lnx = 2𝑣 −1 2 - ln v
  • 5. SOLUTION : ln x + ln v = 2𝑣 −1 2 ln (x+v) = 2 ( 𝑦 𝑥 ) −1 2 ln [x+ 𝑦 𝑥 ] =2 ( 𝑦 𝑥 ) −1 2 (Ans)