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

\dfrac{\partial M(x,y)}{\partial y} \neq \dfrac{\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

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

u \dfrac{\partial M}{\partial y} + M \dfrac{\partial u}{\partial y} = u ~\dfrac{\partial N}{\partial x} + N \dfrac{\partial u}{\partial y}

u \left( \dfrac{\partial M}{\partial y} -\dfrac{\partial N}{\partial x} \right) = -\left(M \dfrac{\partial u}{\partial y} -N \dfrac{\partial u}{\partial y} \right)

RUMUS UMUM FAKTOR INTEGRAL

u(x, y) = \dfrac{-\left( M ~\frac{\partial u}{\partial y} -N ~\frac{\partial u}{\partial x} \right)}{\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

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

Oleh karena itu, Rumus Umum Faktor Integral diatas dapat ditulis

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

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

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

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

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

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

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

dengan h(x) = \dfrac{\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

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

Oleh karena itu, Rumus Umum Faktor Integral diatas dapat ditulis

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

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

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

\displaystyle \int \dfrac{-\left( \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x} \right)}{M} ~dy = \int \dfrac{du}{u(y)}

\displaystyle \int \dfrac{-\left( \frac{\partial M}{\partial y} -\frac{\partial N}{\partial x} \right)}{M}~ dy = \ln u

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

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

dengan h(y) = \dfrac{\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)

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

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

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

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

u(x, y) = \dfrac{-\left( M ~\frac{\partial u}{\partial y} -N ~\frac{\partial u}{\partial x} \right)}{\frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}}

u(v) = \dfrac{-\left( M \left( \frac{\partial u}{\partial v} ~\frac{\partial v}{\partial y} \right) -N \left( \frac{\partial u}{\partial v} ~\frac{\partial v}{\partial x} \right) \right)}{\frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}}

-\left( \dfrac{\partial M}{\partial y} -\dfrac{\partial N}{\partial x} \right) u(v) = \dfrac{\partial u}{\partial v} \left(M ~\dfrac{\partial v}{\partial y} -N ~\dfrac{\partial v}{\partial x} \right)

-\left( \dfrac{\partial M}{\partial y} -\dfrac{\partial N}{\partial x} \right) \partial v = \dfrac{\partial u}{u(v)} \left(M ~\dfrac{\partial v}{\partial y} -N \dfrac{\partial v}{\partial x} \right)

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

\displaystyle \int \dfrac{du}{u} = \int \dfrac{-\left( \frac{\partial M}{\partial y}- \frac{\partial N}{\partial x} \right) dv}{M ~\frac{\partial v}{\partial y} -N \frac{\partial v}{\partial x}}

\ln u = \displaystyle \int \dfrac{-\left( \frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} \right) dv}{M~\frac{\partial v}{\partial y} -N ~\frac{\partial v}{\partial x}}

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

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

Contoh 1.

Tentukan Faktor Integral dan penyelesain PD dibawah ini :

  1. (4 xy + 3y^2 -x) dx + x(x + 2y) dy = 0

    Penyelesaian.

    misal : M(x, y) = 4 xy + 3y^2 -x

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

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

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

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

    \dfrac{1}{N} \left( \dfrac{\partial M}{\partial y} -\dfrac{\partial N}{\partial y} \right) = \dfrac{2(x+2y)}{x(x+2y)}

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

    maka FI adalah e^{\displaystyle \int \frac{2}{x} ~dx} = e^{\ln x^2} = x^2

    sehingga diperoleh PD eksak adalah

    x^2 (4 xy + 3y^2 -x) ~dx + x^3 (x + 2y) ~dy = 0

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

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

    ambil \dfrac{\partial F}{\partial y} = x^2 (4 xy + 3y^2 -x)

    = 4x^3y + 3x^2y^2 -x^3

    F(x, y) = \displaystyle \int^x = (4x^3y + 3x^2y^2 -x^3)+ g(y)

    = x^4y + x^3y^2 -\dfrac{1}{4} x^4 + g(y)

    \dfrac{\partial F}{\partial y} = x^4 + 2x^3y + g'(y)

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

    x^4 + 2x^3y + g'(y) = x^3 (x + 2y)

    x^4 + 2x^3y + g'(y) = x^4 + 2x^3y

    g'(y) = 0

    g(y) = C

    solusi PD : x^4y + x^3y^2 -\dfrac{1}{4}x^4 + C

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

    Penyelesaian.

    misal : M(x, y) = xy + y^2 + y

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

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

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

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

    \dfrac{1}{M} \left( \dfrac{\partial M}{\partial y}-\dfrac{\partial N}{\partial y} \right) = \dfrac{-(x+y+1)}{y(x+y+1))}

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

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

    sehingga diperoleh PD eksak adalah

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

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

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

    ambil \dfrac{\partial F}{\partial y} = y^2(x + y + 1)

    = xy^2 + y^3 + y^2

    F(x, y) = \displaystyle \int^x (xy^2 + y^3 + y^2) dx + g(y)

    = \dfrac{1}{2} x^2y^2 + xy^3 + xy^2 + g(y)

    \dfrac{\partial F}{\partial y} = x^2y + 3xy^2 + 2xy + g'(y)

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

    x^2y + 3xy^2 + 2xy + g'(y) = xy(x + 3y + 2)

    x^2y + 3xy^2 + 2xy + g'(y) = x^2y + 3xy^2 + 2xy

    g'(y) = 0

    g(y) = C

    solusi PD : \dfrac{1}{2}x^2y^2 + xy^3 + xy^2 + C

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

    Penyelesaian.

    misal : M(x, y) = 2x^3y^2 -y

    N(x, y) = 2x^2y^3 -x

    \dfrac{\partial M}{\partial y} = 4x^3y -1

    \dfrac{\partial N}{\partial x} = 4xy^3 -1 

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

    \dfrac{\partial M}{\partial y} -\dfrac{\partial N}{\partial x} = (4x^3y -1) -(4xy^3 -1)

    = 4xy(x^2 -y^2)

    ambil :

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

    M \dfrac{\partial v}{\partial y} = x(2x^3y^2 -y)

    N \dfrac{\partial v}{\partial x} = y(2x^2y^3 -x)

    maka

    M \dfrac{\partial v}{\partial y}- N \dfrac{\partial v}{\partial x}

    = (2x^4y^2 -xy) - (2x^2y^4 -xy)

    = 2x^2y^2(x^2 -y^2)

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

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

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

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

    = e^{-2 \ln xy}

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

    sehingga diperoleh PD eksak adalah

    \dfrac{1}{x^2y^2} (2x^3y^2 -y) dx + \dfrac{1}{x^2y^2} (2x^2y^3 -x) dy = 0

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

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

    ambil \dfrac{\partial F}{\partial y} = \dfrac{1}{x^2y^2} (2x^3y^2 -y)

    = 2x- \dfrac{1}{x^2y}

    F(x, y) = \displaystyle \int^x \left( 2x -\dfrac{1}{x^2y} \right) dx + g(y)

    = x^2 + \dfrac{1}{xy} + g(y)

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

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

    -\dfrac{1}{xy^2} + g'(y) = \dfrac{1}{x^2y^2} (2x^2y^3 -x)

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

    g'(y) = 2y

    g(y) = y^2

    solusi PD : x^2 + \dfrac{1}{xy} + y^2 = 0

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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