Perdif Non Homogen Bentuk Khusus (ax + by + c)dx + (px + qy + r)dy = 0
Bentuk umum Perdif non homogen bentuk khusus:
(ax + by + c)dx + (px + qy + r)dy = 0
terdapat 3 kondisi:
• a/p = b/q = c/r
• a/p = b/q ≠ c/r
• a/p ≠ b/q
1. Perdif Non Homogen Bentuk Khusus dengan a/p = b/q = c/r
Misal suatu Perdif (ax + by + c)dx + (px + qy + r)dy = 0 dengan
a/p = b/q = c/r = λ
Misal px + qy + r = u ↔ ax + by + c = λu, perdif menjadi
λu dx + u dy = 0, bagi masing-masing ruas dengan u
λ dx + dy = 0
∫λ dx + ∫dy = 0
λx + y = C
Jadi, hasil akhirnya adalah λx + y = C.
2. Perdif Non Homogen Bentuk Khusus dengan a/p = b/q ≠ c/r
Misal suatu Perdif (ax + by + c)dx + (px + qy + r)dy = 0 dengan
a/p = b/q ≠ c/r
Misal a/p = b/q = μ; px + qy = u → p dx + q dy = du ↔ dy = (du − p dx)/q
perdif menjadi
(μu + c)dx + (u + r)(du − p dx)/q = 0, kalikan q
(μuq + cq − pu − pr)dx + (u + r)du = 0
dx = −(u + r)/(μuq + cq − pu − pr) du
bentuk terakhir ini separabel.
3. Perdif Non Homogen Bentuk Khusus dengan a/p ≠ b/q
Misal suatu Perdif (ax + by + c)dx + (px + qy + r)dy = 0 dengan
a/p ≠ b/q
Misal ax + by + c = u, px + qy + r = v
a dx + by = du ...(i)
p dx + qy = dv ...(ii)
selesaikan SPL ini dan diperoleh
dx = [q du − b dv]/(aq − bp), dy = [a dv − p du]/(aq − bp)
perdif menjadi
u[q du − b dv]/(aq − bp) + v[a dv − p du]/(aq − bp) = 0, kalikan aq − bp
u[q du − b dv] + v[a dv − p du] = 0
(uq − vp)du + (av − bu)dv = 0
bentuk terakhir ini homogen.
Contoh Soal
1. (2x – 5y + 2)dx + (10y – 4x – 4)dy = 0
(2x – 5y + 2)dx + (–4x + 10y – 4)dy = 0
Perhatikan
2/(–4) = –5/10 = 2/(–4) = –½
solusi:
y = ½x + C
2. (3x + 2y + 1)dx – (3x + 2y – 1)dy = 0
(3x + 2y + 1)dx + (–3x – 2y + 1)dy = 0
Perhatikan
3/(–3) = 2/(–2) ≠ 1/1
Misal u = –3x – 2y → du = –3dx – 2dy ↔ dy = (du + 3 dx)/(–2)
perdif menjadi
(–u + 1)dx + (u + 1)(du + 3 dx)/(–2) = 0, kalikan 2
(–2u + 2 – 3u – 3)dx + (–u – 1)du = 0
dx = (–u – 1)/(5u + 1) du = [–⅕ – ⅘/(5u + 1)]du
(25/2)x + (5/2)u + 2⋅ln|5u + 1| = C₄
(25/2)x + (5/2)(–3x – 2y) + 2⋅ln|5(–3x – 2y) + 1| = C₄
5x – 5y + 2⋅ln|–15x – 10y + 1| = C₄
3. dy/dx = (1 – 2y – 4x)/(1 + y + 2x)
(2x + y + 1)dy = (–4x – 2y + 1)dx
(4x + 2y – 1)dx + (2x + y + 1)dy = 0
Perhatikan
4/2 = 2/1 ≠ –1/1
Misal u = 2x + y → du = 2 dx + dy ↔ dy = du – 2 dx
perdif menjadi
(2u – 1)dx + (u + 1)(du – 2 dx) = 0
(2u – 1 – 2u – 2)dx + (u + 1)du = 0
(–3)dx + (u + 1)du = 0
dx = (u + 1)/3 du = (u/3 + 1/3)du
∫dx = ∫u/3 + 1/3 du
x + C₁ = u²/6 + u/3 + C₂, kalikan 6
6x = u² + 2u + C₃
6x = (2x + y)² + 2(2x + y) + C₃ = 4x² + 4xy + y² + 4x + 2y + C₃
4x² + 4xy + y² – 2x + 2y = C₄
4. (2x – 5y + 3)dx – (2x + 4y – 6)dy = 0
(2x – 5y + 3)dx + (–2x – 4y + 6)dy = 0
Perhatikan
2/(–2) ≠ (–5)/(–4)
Misal u = 2x – 5y + 3, v = –2x – 4y + 6
(−4u + 2v)du + (2v + 5u)dv = 0
Misal v = tu → dv = u dt + t du
(−4u + 2tu)du + (2tu + 5u)(u dt + t du) = 0
(−4u + 2tu + 2t²u + 5tu)du + (2tu² + 5u²)dt = 0, bagi dengan u
(−4 + 7t + 2t²)du + u(2t + 5)dt = 0
(−4 + 7t + 2t²)du = −u(2t + 5)dt
1/u du = (2t + 5)/(4 − 7t − 2t²) dt
ln|u| + C₁ = −⅔⋅ln|1 − 2t| − ⅓⋅ln|4 + t| + C₂, kalikan 3
3⋅ln|u| = −2⋅ln|1 − 2t| − ln|4 + t| + C₃, eksponensialkan masing-masing ruas
|u|³ = C₄⋅(1 − 2t)⁻²⋅|4 + t|⁻¹, kembalikan t = v/u|u|³⋅(1 − 2v/u)²⋅|4 + v/u| = C₄
(u − 2v)²⋅|4u + v| = C₄
[2x − 5y + 3 − 2(−2x − 4y + 6)]²⋅|4(2x − 5y + 3) + (−2x − 4y + 6)| = C₄
(6x + 3y − 9)²⋅|6x − 24y + 18| = C₄
5. (3y – 7x + 7)dx + (7y – 3x + 3)dy = 0
(–7x + 3y + 7)dx + (–3x + 7y + 3)dy = 0
Perhatikan
(–7)/(–3) ≠ 3/7
Misal u = –7x + 3y + 7, v = –3x + 7y + 3
(uq − vp)du + (av − bu)dv = 0
(7u + 3v)du + (–7v – 3u)dv = 0
Misal v = tu → dv = u dt + t du
(7u + 3tu)du + (–7tu – 3u)(u dt + t du) = 0
(7u + 3tu – 7t²u – 3tu)du + (–7tu² – 3u²)dt = 0, bagi dengan u
(7 – 7t²)du + u(–7t – 3)dt = 0
(7 – 7t²)du = –u(–7t – 3)dt
1/u du = (7t + 3)/(7 – 7t²) dt
7⋅ln|u| + 2⋅ln|1 + t| + 5⋅ln|1 – t| = C₃, eksponensialkan masing-masing ruas
|u|⁷⋅(1 + t)²⋅|1 – t|⁵ = C₄
(u + tu)²⋅|u – tu|⁵ = C₄, kembalikan v = tu
(u + v)²⋅|u – v|⁵ = C₄
(–7x + 3y + 7 + (–3x + 7y + 3))²⋅|–7x + 3y + 7 – (–3x + 7y + 3)|⁵ = C₄
(–10x + 10y + 10)²⋅|–4x – 4y + 4|⁵ = C₄
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