Suatu persamaan dierensial exsak dengan bentuk :
\(M \left ( x,y \right ) dx + N\left ( x,y \right )dy= 0\)
Dikatakan persamaan diferenisal eksak, jika ada suatu fungsi \(f\left ( x,y \right )\) yang diferensial totalnya sama dengan \(M \left ( x,y \right ) dx + N\left ( x,y \right )dy\), yaitu (dengan meniadakan lambang x dan y):
\(df= M dx + N dy\)
uji kepastian: jika M dan N merupakan fungsi kontinu dan memiliki turunan parsial pertama yang kontinu pada sebuah segiempat bidang xy, maka \(M \left ( x,y \right ) dx + N\left ( x,y \right )dy= 0\) adalah eksak hanya jika \(\frac{\partial M}{\partial y}= \frac{\partial N}{\partial x}\)
Metode Solusi
Untuk menentukan solusi dari \(M \left ( x,y \right ) dx + N\left ( x,y \right )dy= 0\), maka secara implisit diberikan oleh penyelesaian umum \(f\left ( x,y \right ) = c \).
Langkah-langkan menentukan suatu fungsi \(f\left ( x,y \right )\) adalah:
Langkah 1
perhatikan bahwa :
\(\frac{\partial f}{\partial x}= M\left ( x,y \right ), dan \frac{\partial f}{\partial y}= N\left ( x,y \right )\)
Langkah 2
integrasikan (mencari integral) \(M\left ( x,y \right )\) terhadap x dengan y tetap
\(\frac{\partial f}{\partial x}dx = M\left ( x,y \right )dx\)
\(f\left ( x,y \right )=\left [ \int ^{x}M\left ( x,y \right )dx \right ] + \varnothing \left ( y \right )\)
dimana \(\varnothing y\) adalah fungsi sembarang dari y saja.
Langkah 3
fungsi \(f\left ( x,y \right )\) pada langkah ke 2, dideferensialkan parsial terhadap y yang selanjutnya akan diperoleh :
\(\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left [ \int ^{x} M\left ( x,y \right ) dx \right ]+\frac{\partial \varnothing }{\partial y}\)\
Langkah 4
karena \(\frac{\partial f}{\partial y}=N\left ( x,y \right )\) maka,
\(\frac{\partial \varnothing }{\partial y}= N\left ( x,y \right ) - \frac{\partial }{\partial y}\left [ \int ^{x} M\left ( x,y \right )dx \right ]\)
Langkah 5
\(\varnothing\) \(\left ( y \right )\) yang baru saja diperoleh, disubstitusikan ke \(f\left ( x,y \right )\) dalam langkah ke 2. dengan demikian \(f\left ( x,y \right ) = C\) dapat diperoleh.
Contoh Soal :
1. Buktikan bahwa persamaan \(\left ( 2x+3y \right )dx + \left ( 3x+4y \right )dy=0\) adalah eksak dan selesaikan!
Jawab :
\(P = 2x + 3y \rightarrow \frac{\partial P}{\partial y}=3\)
\(Q=3x+4y\rightarrow \frac{\partial Q}{\partial x}=3\)
\(\frac{\partial f}{\partial x} = P\)
\(\frac{\partial f}{\partial y}=Q\)
\(\frac{\partial P}{\partial y}=\frac{\partial q}{\partial x}\) merupakan PDB Eksak
\(P=\frac{\partial f}{\partial x}\)
\(f\left ( x,y \right )=\int \left ( 2x+3y \right )dx=x^{2}+3yx+c\left ( y \right )\)
\(Q=\frac{\partial f}{\partial y}=3x+4y)\
3x+{c}'\left ( y \right )\)=3x+4y
\({c}'\left ( y \right )=4y\)
\(c\left ( y \right )=\int 4y \ dy\)
\(c\left ( y \right )=2y^{2}\)
jadi \(f\left ( x,y \right )=x^{2}+3xy+2y^{2}\)
2. Buktikan bahwa persamaan \(sin \left ( x+y \right )dx+\left ( 5y^{2}+3y+sin\left ( x+y \right ) \right )dy=0\) adalah eksak dan selesaikan!
Jawab :
\(M=sin\left ( x+y \right )\)
\(N=\left ( 5y^{2}+3y+sin\left ( x+y \right ) \right )\)
\(\frac{\partial M}{\partial y}=cos\left ( x+y \right )\)
\(\frac{\partial N}{\partial x}=cos\left ( x+y \right )\)
\(u=\int Mdx+k\left ( y \right )=\int cos\left ( x+y \right )dx+k\left ( y \right )\)
\( = cos\left ( x+y \right ) +k\left ( y \right )\)
mencari nilai k\(\left ( y \right )\)
\(\frac{\partial u}{\partial y}= -sin\left ( x+y \right )+\frac{\partial k}{\partial y}= N=5y^{2}+3y+sin\left ( x+y \right )\)
\(\frac{dk}{dy}=5y^{2}-3y\)
\(k\left ( y \right )=\frac{5}{3}y^{3}+\frac{2}{3}y^{2}+C\)
jadi
\(u\left ( x,y \right )=sin\left ( x+y \right )+\frac{5}{3}y^{3}+\frac{3}{2}y^{2}+c\)
3. Buktikan bahwa persamaan diferensial : \(\left ( x+y \right )dx + \left ( x-y \right )dy+0\), adalah eksak dan selesaikan!
Pembuktian:
\(M\left ( x,y \right )=x+y\rightarrow \frac{\partial M\left ( x,y \right )}{\partial y}=1 \)
\(N\left ( x,y \right )=x-y\rightarrow \frac{\partial N\left ( x,y \right )}{\partial x}=1\)
Maka: \(\left ( x+y \right )dx + \left ( x-y \right )dy=0\) adalah persamaan eksak
Penyelesaian:
Menentukan penyelesaian dengan persamaan \(M\left ( x,y \right )=\frac{\partial F\left ( x,y \right )}{\partial x}\)
\(M\left ( x,y \right )\frac{\partial F\left ( x,y \right )}{\partial x}=x+y\)
\(\int \partial F\left ( x,y \right )=\int \left ( x+y \right )\partial x\)
\(F\left ( x,y \right )\int ^{x}\left ( x+y \right )\partial x+g\left ( y \right )=\frac{x^{2}}{2}+xy+g\left ( y \right )\)
\(\frac{\partial F\left ( x,y \right )}{\partial y}=\frac{\partial }{\partial y}\left [ \frac{x^{2}}{2}+xy \right ]+\frac{dg\left ( y \right )}{dy}=x+\frac{dg\left ( y \right )}{dy}\)
\(x-y=x+\frac{dg\left ( y \right )}{dy}\rightarrow \frac{dg\left ( y \right )}{dy}=-y\rightarrow dg\left ( y \right )=-yd\)
\(\int dg\left ( y \right )=\int -yd\rightarrow g\left ( y \right )=-\frac{y^{2}}{2}\)
jadi diperoleh penyelesaian:
\(F\left ( x,y \right )=\frac{x^{2}}{2}+xy-\frac{y^{2}}{2}=C\)
Bentuk umum persamaan diferensial biasa orde pertama:
\(\frac{dy}{dx}\)=\(F(x,y)\)
Dalam beberapa kasus fungsi \(f(x,y)\) di sebelah
kanan persamaan dapat dituliskan :
\(f(x,y)\)=\(-\frac{M(x,y)}{N(x,y)}\)
\(\frac{dy}{dx}\)=\(-\frac{M(x,y)}{N(x,y)}
\(M(x,y)dx+N(x,y)dy\)=\(0\)
Definisi Persamaan Eksak
\(M(x,y)dx+N(x,y)dy\)=\(0\)
disebut persamaan eksak jika ada fungsi
kontinue di \(u(x,y)\)
\(du\)=\(M(x,y)dx+N(x,y)dy\)=\(0\)
contoh soal :
1. \((x^{2}+y)dx+(2y^{2}+x)dy\) =\(0\)
\(F(x,y)\)=\(\int M(x.y)dx+Q(y)\)
=\(\int (x^{2}+y)dx+Q(y)\)
=\(\frac{1}{3}x^{3}+xy+Q(y)\)
=\(\frac{\partial F}{\partial y}\)=\(x+Q'(y)
\(2y^{2}+x\)=\(x+Q'(y)\)
\(Q'(y)\) =\(2y^{2}\)=\(\frac{2}{3}y^{2}\)
\(F(x,y)\)=\(\frac{1}{3}x^{3}+xy+\frac{2}{3}y^{3}\)
2.\((x+2y)dx+(4y+2x)dy\)=\(0\)
\(F(x,y)\)=\(\int N(x.y)dy+Q)(x)\)
\(\int(4y+2x)dy+Q(y)\)=\(2y^{2}+2xy+Q(x)\)
\(\frac{\partial F}{\partial x}\)=\(2y+Q'(x)\)
\(x+2y\) =\(2y+Q'(x)\)
\(Q'(x)\)=\(x\)
\(Q(x)\) =\(\int x dx\)=\(\frac{1}{2}x^{2}\)
\(F(x,y)\)=\(\frac{1}{2}x^{2}+2xy+2y^{2}\)
\(F(x,y)\)=\(\int N(x.y)dx+Q(x)\)=\(\int(x+2y)dx+Q(y)\)
=\(\frac{1}{2}x^{2}+2xy+Q(y)\)
\(\frac{\partial F}{\partial y}\)=\(2x+Q'(y)\)
\(4x+2y\)=\(2x+Q'(y)\)
\(Q'(y)\)=\(4y\)
\(Q(y)\) =\(\int 4y dy\) =\(2y^{2}\)
\(F(x,y)\)=\(\frac{1}{2}x^{2}+2xy+2y^{2}\)
3.\((2x+2y)dx+(y^{3}+2x)dy\)=\(0\)
\(F(x,y)\)=\(\int M(x.y)dx+Q(y)\)
=\((2x+2y)dx+Q(y)\)
=\(x^{2}+2xy+Q(y)\)
\(\frac{\partial F}{\partial y}\)=\(2x+Q'(y)\)
\(y^{3}+2x\)=\(2x+Q'(y)\)
\(Q'(y)\)=\(y^{3}\)
\(Q(y)\)=\(\int y^{3}dy\)=\(\frac{1}{4}y^{4}\)
\(F(x,y)\)=\(x^{2}+2xy+\frac{1}{4}x^{4}\)
\(F(x,y)\)=\(\int N(x.y)dy+Q(x)\)
=\(\int (y^{3}+2x)dy+Q(x)\)
=\(\frac{1}{4}y^{4}+2xy+Q(x)\)
\(\frac{\partial F}{\partial x}\)=\(2x+Q'(x)\)
\(2x+2y\)=\(2y+Q'(x)\)
\(Q'(x)\)=\(2x\)
\(Q(x)\)=\(\int 2x dx\)=\(x^{3}\)
\(F(x,y)\)=\(x^{2}+2xy+\frac{1}{4}y^{3}\)
Persamaan diferensial eksak
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