Persamaan Diferensial Eksak

Misal suatu persamaan diferensial

M(x, y) dx + N(x, y) dy = 0

persamaan diferensial dalam bentuk ini dikatakan eksak jika berlaku

langkah-langkah menyelesaikan Perdif eksak:

1. Integralkan M(x, y) terhadap x atau N(x, y) terhadap y

misal dipilih mengintegralkan M(x, y) terhadap x. F(x, y) = ∫M(x, y) dx + h(y)

2. Turunkan parsial F(x, y) terhadap y, lalu samakan dengan N(x, y)

3. Integralkan h'(y) untuk memperoleh h(y)

4. Penyelesaiannya adalah F(x, y) = C

Contoh Soal Cek Eksak

1. (y² – 2x)dx + (2x + 1)dy = 0

∂M/∂y = 2y ≠ ∂N/∂x = 2, bukan Perdif eksak

2. (3x²y – 1)dx + (x³ + 6y – y²)dy = 0; y(0) = 3

∂M/∂y = 3x² = ∂N/∂x, Perdif eksak

3. (2xy – 9x²)dx + (2y + x² + 1)dy = 0; y(0) = –3

∂M/∂y = 2x = ∂N/∂x = 2x, Perdif eksak

4. (6x + y²)dx + (2xy – 3y²)dy = 0

∂M/∂y = 2y = ∂N/∂x, Perdif eksak

5. [cos(2y) – 3x²y²]dx + [cos(2y) – 2x.sin(2y) – 2x³y]dy = 0

∂M/∂y = –2sin(2y) – 6x²y = ∂N/∂x, Perdif eksak

6. [r + sin(θ) – cos(θ)]dr + [r.sin(θ) + r.cos(θ)]dθ = 0

M(r, θ) = r + sin(θ) – cos(θ); N(r, θ) = r.sin(θ) + r.cos(θ)

∂M/∂θ = cos(θ) + sin(θ) = ∂N/∂x, Perdif eksak

7. [sin(θ) – 2r².cos(θ)]dr + r.cos(θ).[2r.sin(θ) + 1]dθ = 0

∂M/∂θ = cos(θ) + 2r².sin(θ) ≠ ∂N/∂r = 4r.sin(θ).cos(θ) + cos(θ), bukan Perdif eksak

8. (2xy)dx + (x² + y²)dy = 0

∂M/∂y = 2x = ∂N/∂x, Perdif eksak

9. [y³ + cos(x + y²)]dx + [3xy² + 2y.cos(x + y²)]dy = 0

∂M/∂y = 3y² – 2y.sin(x + y²) = ∂N/∂x, Perdif eksak

11. (2x²y + 4y)y' + 2xy² + 2 = 0

(2x²y + 4y)dy + (2xy² + 2)dx = 0

(2xy² + 2)dx + (2x²y + 4y)dy = 0

∂M/∂y = 4xy = ∂N/∂x, Perdif eksak

12. (3x² + 2xy²)dx + (2x²y + 6y²)dy = 0; y(1) = 2

∂M/∂y = 4xy = ∂N/∂x, Perdif eksak

13. (e^2xy² – 2x)dx + (e^2xy)dy = 0; y(0) = 2

∂M/∂y = 2e^2xy = ∂N/∂x, Perdif eksak

Solusi untuk Perdif eksak yang sudah dicek

2. (3x²y – 1)dx + (x³ + 6y – y²)dy = 0; y(0) = 3

F(x, y) = ∫M(x, y) dx + h(y) = ∫(3x²y – 1)dx + h(y) = x³y – x + h(y)

∂F/∂y = x³ + h'(y) = N(x, y) = x³ + 6y – y²

h'(y) = 6y – y²

h(y) = ∫6y – y² dy = y² – ⅓y³

F(x, y) = x³y – x + y² – ⅓y³ = C, masukkan y(0) = 3

C = 0³.3 – 0 + 3² – ⅓.3³ = 0

solusi akhir adalah x³y – x + y² – ⅓y³ = 0

3. (2xy – 9x²)dx + (2y + x² + 1)dy = 0; y(0) = –3

F(x, y) = ∫M(x, y) dx + h(y) = ∫(2xy – 9x²)dx + h(y) = x²y – 3x³ + h(y)

∂F/∂y = x² + h'(y) = N(x, y) = 2y + x² + 1

h'(y) = 2y + 1

h(y) = ∫2y + 1 dy = y² + y

F(x, y) = x²y – 3x³ + y² + y = C, masukkan y(0) = –3

C = 0²(–3) – 3.0³ + (–3)² + (–3) = 6

x²y – 3x³ + y² + y = 6

4. (6x + y²)dx + (2xy – 3y²)dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫(6x + y²)dx + h(y) = 3x² + xy² + h(y)

∂F/∂y = 2xy + h'(y) = N(x, y) = 2xy – 3y²

h'(y) = –3y²

h(y) = –y³

F(x, y) = 3x² + xy² – y³ = C

3x² + xy² – y³ = C

5. [cos(2y) – 3x²y²]dx + [cos(2y) – 2x.sin(2y) – 2x³y]dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫[cos(2y) – 3x²y²]dx + h(y) = x.cos(2y) – x³y² + h(y)

∂F/∂y = –2x.sin(2y) – 2x³y + h'(y) = N(x, y) = cos(2y) – 2x.sin(2y) – 2x³y

h'(y) = cos(2y)

h(y) = ½sin(2y)

F(x, y) = x.cos(2y) – x³y² + ½sin(2y) = C

x.cos(2y) – x³y² + ½sin(2y) = C

6. [r + sin(θ) – cos(θ)]dr + [r.sin(θ) + r.cos(θ)]dθ = 0

F(r, θ) = ∫M(r, θ) dr + h(θ) = ∫[r + sin(θ) – cos(θ)]dr + h(θ) = ½r² + r.sin(θ) – r.cos(θ) + h(θ)

∂F/∂θ = r.cos(θ) + r.sin(θ) + h'(θ) = N(r, θ) = r.sin(θ) + r.cos(θ)

h'(θ) = 0

h(θ) = C₁

F(r, θ) = ½r² + r.sin(θ) – r.cos(θ) + C₁ = C₂

½r² + r.sin(θ) – r.cos(θ) = C₃; dengan C₃ = C₂ – C₁

8. (2xy)dx + (x² + y²)dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫(2xy)dx = x²y + h(y)

∂F/∂y = x² + h'(y) = N(x, y) = x² + y²

h'(y) = y²

h(y) = ⅓y³

F(x, y) = x²y + ⅓y³ = C

x²y + ⅓y³ = C

9. [y³ + cos(x + y²)]dx + [3xy² + 2y.cos(x + y²)]dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫[y³ + cos(x + y²)]dx = xy³ + sin(x + y²) + h(y)

∂F/∂y = 3xy² + 2y.cos(x + y²) + h'(y) = N(x, y) = 3xy² + 2y.cos(x + y²)

h'(y) = 0

h(y) = C₁

F(x, y) = 3xy² + 2y.cos(x + y²) + C₁ = C₂

3xy² + 2y.cos(x + y²) = C₃; dengan C₃ = C₂ – C₁

10. (ye^2xy + x)dx + (bxe^2xy)dy = 0

tentukan b agar Perdif ini eksak, lalu selesaikan.

∂M/∂y = e^2xy + 2xye^2xy 

∂M/∂x = be^2xy + 2xye^2xy 

agar Perdif ini eksak, diharuskan b = 1.

(ye^2xy + x)dx + (xe^2xy)dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫(ye^2xy + x)dx = e^2xy/2 + x²/2 + h(y)

∂F/∂y = xe^2xy + h'(y) = N(x, y) = xe^2xy 

h'(y) = 0

h(y) = C₁

F(x, y) = xe^2xy + C₁ = C₂

xe^2xy = C₃; dengan C₃ = C₂ – C₁

11. (2x²y + 4y)y' + 2xy² + 2 = 0

(2xy² + 2)dx + (2x²y + 4y)dy = 0

F(x, y) = ∫M(x, y) dx + h(y) = ∫(2xy² + 2)dx = x²y² + 2x + h(y)

∂F/∂y = 2x²y + h'(y) = N(x, y) = 2x²y + 4y

h'(y) = 4y

h(y) = 2y²

F(x, y) = x²y + ⅓y³ = C

x²y + ⅓y³ = C

12. (3x² + 2xy²)dx + (2x²y + 6y²)dy = 0; y(1) = 2

F(x, y) = ∫M(x, y) dx + h(y) = ∫(3x² + 2xy²)dx + h(y) = x³ + x²y² + h(y)

∂F/∂y = 2x²y + h'(y) = N(x, y) = 2x²y + 6y²

h'(y) = 6y²

h(y) = 2y³

F(x, y) = x³ + x²y² + 2y³ = C, masukkan y(1) = 2

C = 1³ + 1².2² + 2.2³ = 21

x³ + x²y² + 2y³ = 21

13. (e^2xy² – 2x)dx + (e^2xy)dy = 0; y(0) = 2

F(x, y) = ∫M(x, y) dx + h(y) = ∫(e^2xy² – 2x)dx + h(y) = ½e^2xy² – x² + h(y)

∂F/∂y = e^2xy + h'(y) = N(x, y) = e^2xy

h'(y) = 0

h(y) = C₁ = C₂

F(x, y) = ½e^2xy² – x² + C₁ = C₂

½e^2xy² – x² = C₃, masukkan y(0) = 2

C₃ = ½e^2.0.2² – 0² = 2

½e^2xy² – x² = 2