Penyelesaian Persamaan Diferensial : Eksak dan Tak Eksak

PD Eksak

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

M(x, y) dx + N(x, y) dy = 0 … (i)

serta jika memenuhi

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

Contoh :

y dx + x dy = 0

misal : M(x, y) = y $\Rightarrow$ $\frac{\partial M(x,y)}{\partial y}$ = 1

N(x, y) = x $\Rightarrow$ $\frac{\partial N(x,y)}{\partial x}$ = 1

karena $\frac{\partial M(x,y)}{\partial y}$ = $\frac{\partial N(x,y)}{\partial x}$ , maka PD diatas merupakan PD eksak.

(2xy + ln x) dx + x2 dy = 0

misal : M(x, y) = 2xy + ln x $\Rightarrow$ $\frac{\partial M(x,y)}{\partial y}$ = 2x

N(x, y) = x2 $\Rightarrow$ $\frac{\partial N(x,y)}{\partial x}$ = 2x

karena $\frac{\partial M(x,y)}{\partial y}$ = $\frac{\partial N(x,y)}{\partial x}$ , maka PD diatas merupakan PD eksak.

(x – y) dx + (x + y) dy = 0

misal : M(x, y) = x – y $\Rightarrow$ $\frac{\partial M(x,y)}{\partial y}$ = -1

N(x, y) = x + y $\Rightarrow$ $\frac{\partial N(x,y)}{\partial x}$ = 1

karena $\frac{\partial M(x,y)}{\partial y}$ $\neq$ $\frac{\partial N(x,y)}{\partial x}$ , maka PD diatas bukan merupakan PD eksak.

Jika F adalah suatu fungsi dua peubah yang mempunyai derivative parsial tingkat satu yang kontinyu dalam suatu domain D, maka diferensial total fungsi F yaitu dF didefinisikan oleh

dF(x) = $\frac{\partial F(x,y)}{\partial x}$ dx + $\frac{\partial F(x,y)}{\partial y}$ dy, $\forall$ (x, y) $\epsilon$ D

Misal penyelesain umum PD (i) adalah F(x, y) = C dengan C adalah konstanta sebarang, maka

dF(x, y) = 0, sedemikian sehingga

$\frac{\partial F(x,y)}{\partial x}$ dx + $\frac{\partial F(x,y)}{\partial y}$ dy = 0 … (ii)

dari PD (i) dan pers (ii), diperoleh

(a) $\frac{\partial F(x,y)}{\partial x}$ = M(x, y)

(b) $\frac{\partial F(x,y)}{\partial y}$ = N(x, y)

Sehingga solusi PD Eksak berbentuk F(x, y) = C. Berdasarkan hal tersebut, dapat dicari solusi PD sebagai berikut :

(a) $\frac{\partial F(x,y)}{\partial x}$ = M(x, y)

F(x, y) = $\int^x$ M(x, y) dx + g(y)

NOTE : bentuk $\int^x$ adalah integral terhadap x, dimana y dipandang sebagai konstanta dan g(y) konstanta integral yang harus dicari.

$\frac{\partial F(x,y)}{\partial y}$ = $\frac{\partial}{\partial y}$ [ $\int^x$ M(x, y) dx] + g'(y)

Karena $\frac{\partial F(x,y)}{\partial y}$ = N(x, y) maka

$\frac{\partial F(x,y)}{\partial y}$ = $\frac{\partial}{\partial y}$ [ $\int^x$ M(x, y) dx] + g'(y) = N(x, y)

g'(y) = N(x, y) – $\frac{\partial}{\partial y}$ [ $\int^x$ M(x, y) dx]

karena g'(y) merupakan fungsi dengan peubah y saja maka setelah disederhanakan merupakan fungsi dari y atau konstanta. Dengan kata lain g(y) dapat dicari

(b) $\frac{\partial F(x,y)}{\partial y}$ = N(x, y)

Integralkan kedua ruas terhadap variabel y, diperoleh

F(x, y) = $\int^y$ N(x, y) dy + f(x)

turunkan kedua ruas dengan turunan parsial terhadap x

$\frac{\partial F(x,y)}{\partial x}$ = $\frac{\partial}{\partial x}$ [ $\int^y$ N(x, y) dy] + f'(x)

karena $\frac{\partial F(x,y)}{\partial x}$ = M(x, y) maka

$\frac{\partial F(x,y)}{\partial x}$ = $\frac{\partial}{\partial x}$ [ $\int^y$ N(x, y) dy] + f'(x) = M(x, y)

f'(x) = M(x, y) – $\frac{\partial}{\partial x}$ [ $\int^y$ N(x, y) dy]

Contoh :

Cari solusi dari PD (x + y) dx + (x – y) dy = 0

Penyelesaian :

Cek terlebih dahulu apakah PD diatas adalah PD eksak.

misal : M(x , y) = x + y $\Rightarrow$ $\frac{\partial M(x,y)}{\partial y}$ = 1

N(x , y) = x – y $\Rightarrow$ $\frac{\partial M(x,y)}{\partial x}$ = 1

karena $\frac{\partial M(x,y)}{\partial y}$ = $\frac{\partial N(x,y)}{\partial x}$ , maka PD tesebut adalah PD eksak.

Untuk mencari solusinya, kita akan menggunakan

F(x, y) = $\int^x$ M(x, y) dx + g(y)

= $\int^x$ (x + y) dx + g(y)

= $\frac{1}{2}$ x2 + xy + g(y)

cari g'(y)

$\frac{\partial F(x,y)}{\partial y}$ = $\frac{\partial}{\partial y}$ [ $\int^x$ M(x, y) dx] + g'(y)

= $\frac{\partial}{\partial y}$ [ $\frac{1}{2}$ x2 + xy] + g'(y)

= x + g'(y)

karena $\frac{\partial F(x,y)}{\partial y}$ = N(x, y), maka

x + g'(y) = N(x, y)

x + g'(y) = x – y

g'(y) = -y

$\int$ g'(y) = $\int$ -y

g(y) = $-\frac{1}{2}$ y2

jadi solusi umumnya : $\frac{1}{2}$ x2 + xy – $\frac{1}{2}$ y2 = c1

x2 + 2xy – y2 = C, dengan C = 2c1

PD : xy’ + y + 4 = 0

Penyelesaian :

x $\frac{dy}{dx}$ + y + 4 = 0

x dy + (y + 4) dx = 0

misal : M(x , y) = y + 4 $\Rightarrow$ $\frac{\partial M(x,y)}{\partial y}$ = 1

N(x , y) = x $\Rightarrow$ $\frac{\partial M(x,y)}{\partial x}$ = 1

karena $\frac{\partial M(x,y)}{\partial y}$ = $\frac{\partial N(x,y)}{\partial x}$ , maka PD tesebut adalah PD eksak.

Untuk mencari solusinya, kita akan menggunakan

F(x, y) = $\int^y$ N(x, y) dy + g(x)

= $\int^y$ x dy + g(x)

= xy + g(x)

cari g'(x)

$\frac{\partial F(x,y)}{\partial x}$ = $\frac{\partial}{\partial x}$ [ $\int^x$ N(x, y) dy] + g'(x)

= $\frac{\partial}{\partial x}$ [xy] + g'(x)

= y + g'(x)

karena $\frac{\partial F(x,y)}{\partial y}$ = N(x, y), maka

y + g'(x) = M(x, y)

y + g'(x) = y + 4

g'(x) = 4

$\int$ g'(x) = $\int$ 4

g(x) = 4x

jadi solusi umumnya : xy + 4x = C

Sumber : https://aimprof08.wordpress.com/2012/12/23/penyelesaian-persamaan-diferensial-pd-eksak/

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) = $\frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{M}$

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) = $e^{h(v) dv}$

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 :

(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
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
(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