Penyelesaian Persamaan Diferensial : PD Tidak Eksak (Faktor Integral)


Persamaan Diferensial Tidak Eksak adalah suatu PD tingkat satu dan berpangkat satu yang berbentuk

M(x, y) dx + N(x, y) dy = 0 … (i)

dan memenuhi syarat

\frac{\partial M(x,y)}{\partial y} \neq \frac{\partial N(x,y)}{\partial x}

Penyelesaian PD tidak eksak dapat diperoleh dengan dengan mengalikan PD (i) dengan suatu fungsi u yang disebut Faktor Integral (FI), sehingga diperoleh PD eksak yaitu

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

karena PD sudah berbentuk eksak, maka memenuhi

\frac{\partial (uM)}{\partial y} = \frac{\partial (uN)}{\partial x}

u \frac{\partial M}{\partial y} + M \frac{\partial u}{\partial y} = u \frac{\partial N}{\partial x} + N \frac{\partial u}{\partial y}

u (\frac{\partial M}{\partial y} \frac{\partial N}{\partial x} ) = – (M \frac{\partial u}{\partial y} – N \frac{\partial u}{\partial y} )

RUMUS UMUM FAKTOR INTEGRAL

u(x, y) = \frac{-(M \frac{\partial u}{\partial y}-N \frac{\partial u}{\partial x})}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}

Secara umum Faktor Integral terdiri dari tiga kasus yaitu

(a) FI u sebagai fungsi x saja

karena u sebagai fungsi x saja, maka

\frac{\partial u}{\partial x} = \frac{du}{dx} dan \frac{\partial u}{\partial y} = 0

Oleh karena itu, Rumus Umum Faktor Integral diatas dapat ditulis

u(x) = \frac{N \frac{du}{dx}}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}

\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} dx = Q \frac{du}{u(x)}

\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{Q} dx = \frac{du}{u(x)}

\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{Q} dx = \int \frac{du}{u(x)}

\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{Q} dx = ln u

u(x) = e^{\int \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} dy}{Q}}

u(x) = e^{\int h(x) dx}

dengan h(x) = \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{Q}

(b) FI u sebagai fungsi y saja

karena u sebagai fungsi y saja, maka

\frac{\partial u}{\partial x} = 0 dan \frac{\partial u}{\partial y} = \frac{du}{dy}

Oleh karena itu, Rumus Umum Faktor Integral diatas dapat ditulis

u(y) = \frac{-M \frac{du}{dy}}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}

\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x} dy = -M \frac{du}{u(y)}

\frac{-(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{M} dy = \frac{du}{u(y)}

\int \frac{-(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{M} dy = \int \frac{du}{u(y)}

\int \frac{-(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x})}{M} dy = ln u

u(y) = e^{-\int (\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) dy}{M}}

u(y) = e^{-\int h(y) dy}

dengan h(y) = \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}

(c) FI u sebagai fungsi x dan y

andaikan FI : u = u(x, y)

misal bentuk peubah x, y = v

maka FI : u = u(v)

\frac{\partial u}{\partial x} = \frac{\partial u}{\partial v} \frac{\partial v}{\partial x} … (iii)

\frac{\partial u}{\partial y} = \frac{\partial u}{\partial v} \frac{\partial v}{\partial y} … (iv)

\frac{\partial u}{\partial y} = \frac{du}{dv} … (v)

Jika pers (iii), (iv) dan (v) disubstitusikan ke RUMUS UMUM FAKTOR INTEGRAL, maka

u(x, y) = \frac{-(M \frac{\partial u}{\partial y}-N \frac{\partial u}{\partial x})}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}

u(v) = \frac{-(M (\frac{\partial u}{\partial v} \frac{\partial v}{\partial y})-N )\frac{\partial u}{\partial v} \frac{\partial v}{\partial x}))}{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}

-(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) u(v) = \frac{\partial u}{\partial v} (M \frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x})

-(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}) \partial v = \frac{\partial u}{u(v)} (M \frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x})

\frac{du}{u} = \frac{-(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}) dv}{M\frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x}}

\int \frac{du}{u} = \int \frac{-(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}) dv}{M\frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x}}

ln u = \int \frac{-(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}) dv}{M\frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x}}

Jadi, FI : u(v) = e^{\int h(v) dv}

dengan h(v) = \frac{-(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}) dv}{M\frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x}}

Contoh :

Tentukan Faktor Integral dan penyelesain PD dibawah ini :

  1. (4 xy + 3y2 – x) dx + x(x + 2y) dy = 0

    Penyelesaian :

    misal : M(x, y) = 4 xy + 3y2 – x

    N(x, y) = x(x + 2y)

    \frac{\partial M}{\partial y} = 4x + 6y

    \frac{\partial N}{\partial x} = 2x + 2y

    Jadi, \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial y}

    \frac{1}{N} (\frac{\partial M}{\partial y}-\frac{\partial N}{\partial y}) = \frac{2(x+2y)}{x(x+2y)}

    = \frac{2}{x} [fungsi dari x saja]

    maka FI adalah e^{\int \frac{2}{x} dx} = e^{ln \quad x^2} = x2

    sehingga diperoleh PD eksak adalah

    x2 (4 xy + 3y2 – x) dx + x3 (x + 2y) dy = 0

    \frac{\partial F}{\partial y} dx + \frac{\partial G}{\partial x} dy = 0

    Karena PD diatas sudah berbentuk PD eksak, sehingga untuk mencari solusinya digunakan Penyelesaian PD Eksak.

    ambil \frac{\partial F}{\partial y} = x2 (4 xy + 3y2 – x)

    = 4x3y + 3x2y2 – x3

    F(x, y) = \int^x (4x3y + 3x2y2 – x3) dx + g(y)

    = x4y + x3y2\frac{1}{4} x4 + g(y)

    \frac{\partial F}{\partial y} = x4 + 2x3y + g'(y)

    karena \frac{\partial F}{\partial y} = G(x, y), sehingga

    x4 + 2x3y + g'(y) = x3 (x + 2y)

    x4 + 2x3y + g'(y) = x4 + 2x3y

    g'(y) = 0

    g(y) = C

    solusi PD : x4y + x3y2\frac{1}{4} x4 + C

  2. y(x + y + 1) dx + x(x + 3y + 2) dy = 0

    Penyelesaian :

    misal : M(x, y) = xy + y2 + y

    N(x, y) = x2 + 3xy + 2x

    \frac{\partial M}{\partial y} = x + 2y + 1

    \frac{\partial N}{\partial x} = 2x + 3y + 2

    Jadi, \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial y}

    \frac{1}{M} (\frac{\partial M}{\partial y}-\frac{\partial N}{\partial y}) = \frac{-(x+y+1)}{y(x+y+1))}

    = -\frac{1}{y} [fungsi dari y saja]

    maka FI adalah e^{\int -\frac{1}{y} dy} = e^{ln \quad y} = y

    sehingga diperoleh PD eksak adalah

    y2(x + y + 1) dx + xy(x + 3y + 2) dy = 0

    \frac{\partial F}{\partial y} dx + \frac{\partial G}{\partial x} dy = 0

    Karena PD diatas sudah berbentuk PD eksak, sehingga untuk mencari solusinya digunakan Penyelesaian PD Eksak.

    ambil \frac{\partial F}{\partial y} = y2(x + y + 1)

    = xy2 + y3 + y2

    F(x, y) = \int^x (xy2 + y3 + y2) dx + g(y)

    = \frac{1}{2} x2y2 + xy3 + xy2 + g(y)

    \frac{\partial F}{\partial y} = x2y + 3xy2 + 2xy + g'(y)

    karena \frac{\partial F}{\partial y} = G(x, y), sehingga

    x2y + 3xy2 + 2xy + g'(y) = xy(x + 3y + 2)

    x2y + 3xy2 + 2xy + g'(y) = x2y + 3xy2 + 2xy

    g'(y) = 0

    g(y) = C

    solusi PD : \frac{1}{2} x2y2 + xy3 + xy2 + C

  3. (2x3y2 – y) dx + (2x2y3 – x) dy = 0

    Penyelesaian :

    misal : M(x, y) = 2x3y2 – y

    N(x, y) = 2x2y3 – x

    \frac{\partial M}{\partial y} = 4x3y – 1

    \frac{\partial N}{\partial x} = 4xy3 – 1

    Jadi, \frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial y}

    \frac{\partial M}{\partial y} \frac{\partial N}{\partial x} = (4x3y – 1) – (4xy3 – 1)

    = 4xy(x2 – y2)

    ambil :

    v = xy \Rightarrow \frac{\partial v}{\partial x} = y dan \frac{\partial v}{\partial y} = x

    M \frac{\partial v}{\partial y} = x(2x3y2 – y)

    N \frac{\partial v}{\partial x} = y(2x2y3 – x)

    maka

    M \frac{\partial v}{\partial y} – N \frac{\partial v}{\partial x}

    = (2x4y2 – xy) – (2x2y4 – xy)

    = 2x2y2(x2 – y2)

    \frac{du}{u} = \frac{-(\frac{\partial M}{\partial y}- \frac{\partial N}{\partial x}) dv}{M\frac{\partial v}{\partial y}-N \frac{\partial v}{\partial x}}

    = \frac{-4xy(x^2-y^2)}{2x^2y^2(x^2-y^2)} dv

    = -\frac{2}{xy} dv [fungsi x dan y]

    maka FI adalah u(x, y) = e^{\int -\frac{2}{xy} dxy}

    = e^{-2.ln \quad xy}

    = \frac{1}{x^2y^2}

    sehingga diperoleh PD eksak adalah

    \frac{1}{x^2y^2} (2x3y2 – y) dx + \frac{1}{x^2y^2} (2x2y3 – x) dy = 0

    \frac{\partial F}{\partial y} dx + \frac{\partial G}{\partial x} dy = 0

    Karena PD diatas sudah berbentuk PD eksak, sehingga untuk mencari solusinya digunakan Penyelesaian PD Eksak.

    ambil \frac{\partial F}{\partial y} = \frac{1}{x^2y^2} (2x3y2 – y)

    = 2x – \frac{1}{x^2y}

    F(x, y) = \int^x (2x – \frac{1}{x^2y} ) dx + g(y)

    = x2 + \frac{1}{xy} + g(y)

    \frac{\partial F}{\partial y} = -\frac{1}{xy^2} + g'(y)

    karena \frac{\partial F}{\partial y} = G(x, y), sehingga

    -\frac{1}{xy^2} + g'(y) = \frac{1}{x^2y^2} (2x2y3 – x)

    -\frac{1}{xy^2} + g'(y) = 2y – \frac{1}{xy^2}

    g'(y) = 2y

    g(y) = y2

    solusi PD : x2 + \frac{1}{xy} + y2 = 0

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12 comments on “Penyelesaian Persamaan Diferensial : PD Tidak Eksak (Faktor Integral)

  1. Ping-balik: Problem (19) : Persamaan Diferensial | Math IS Beautiful

  2. trimakasih sangat membantu sekali ini kak .. yg kasus 3 itu u(v)=e^h(v) apa u(v)=e pangkat integral h(v) kak? aku bingung yg kasus1& 2 ada integral kok yg kasus 3 gada . (^ adalah pangkat)

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