Persamaan diferensial eksak adalah suatu PD tingkat satu dan berpangkat satu yang berbentuk \(M(x,y)dx+N(x,y)dy=0\) serta jika memenuhi \(\frac{\partial M(x,y)}{\partial y} = \frac{\partial N(x,y)}{\partial x}\)
Persamaan diferensial \(M(x,y)dx+N(x,y)dy=0\) disebut persamaan eksak jika ada fungsi kontinyu \(u(x,y\)
\(du= M(x,y) dx+ N(x,y)dy\)
Untuk mengetahui persamaan pertama adalah persamaan eksak
\(\cdot\) Persamaan \(M(x,y)dx+N(x,y)dy=0\)dikatakan persamaan eksak jika dan hanya jika \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)
\(\cdot \)Buktikan \(M(x,y)dx+N(x,y)dy=0\). Kemudian apakah ada fungsi kontinyu \(u(x,y)\) dengan persamaan diferensial \(du=\frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial y}dy\)
\(\cdot \)Bandingkan dengan persamaan \(\frac{\partial u}{\partial x}= M dan \frac{\partial u}{\partial y}= N\)
\(\cdot \)Persamaan diturunkan terhadap y dan x, maka \(\frac{\partial^{2}u}{\partial y \partial x}=\frac{\partial M}{\partial y}\)dan \(\frac{\partial ^{2}u}{\partial x\partial y}=\frac{\partial N }{\partial x}\)
\(\cdot u(x,y)\) kontinyu jika \(\frac{\partial ^{2}u}{\partial y \partial x}=\frac{\partial ^{2}u}{\partial y \partial x}\)
\(\cdot \)Maka akan didapatkan \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)
\(\cdot \)Sebaliknya jika \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\)
\(\cdot \)Maka ditunjukkan ada fungsi \(u(x,y)\) seperti berikut:
\(\frac{\partial u}{\partial x}= M(x,y)\) dan \(\frac{\partial u}{\partial y}= N(x,y)\)
Intergalkan \(\frac{\partial u}{\partial x}\) terhadap x dengan menganggap y tetap
\(u(x,y)= \int M (x,y)dx + \phi(y)\)
\(\cdot\) dimana \(\phi (y)\) adalah fungsi y, untuk membangun fakta tersebut, maka diferensialkan persamaan tersebut terhadap y,
\(\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}\int M (x,y)dx + {\phi}'(y)\)
\(\cdot\) dengan menyamakan persamaan \(\frac{\partial u}{\partial x}\) maka
\({\phi}'(y)= N(x,y)-\frac{\partial }{\partial }\int M(x,y)dx\)
\(\cdot\) Dengan menganggap bahwa diferensial parsial terhadap x dan y, maka
\(\frac{\partial {\phi}'(y)}{\partial x}=\frac{\partial}{\partial x}\left [ N(x,y)-\frac{\partial }{\partial y\int M(x,y)dx} \right ]\)
\(=\frac{\partial N}{\partial x}-\frac{\partial}{\partial x}\left [ \frac{\partial}{\partial y}\int M(x,y)dx \right ]\)
\(=\frac{\partial N}{\partial x}-\frac{\partial }{\partial y}\left [ \frac{\partial}{\partial y}\int M(x,y)dx \right ]\)
\(=\frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\)
\(=0\)
Contoh 1:
\(1.(2x+3y)dx + (3x+4y)dy=0\)
\(P=2x+3y=\frac{\partial f}{\partial x}\)
\(Q=3x+4y=\frac{\partial f}{\partial y}\)
\(\frac{\partial P}{\partial y}=2x+3y=3\)
\(\frac{\partial Q}{\partial X}=3x+4y=3\)
merupakan PD eksak karna \(\frac{\partial p}{\partial y}=\frac{\partial Q}{\partial x}\)
Mencari P
\(P=\frac{\partial f}{\partial x}\)
\(f=(x,y)\)
\(=\int2x+3y dx\)
\(=x^{2}+3y+c(y)\)
\(Q=\frac{\partial f}{\partial y}=3x+4y\)
\(\partial f = x^{2}+3yx+c(y)\)
\(\frac{\partial f}{\partial y}= 3x+{c}'(y)=3x+4y\)
\({c}'(y)=4y\)
\(c(y)=\int4y dy\)
\(cy=2y^{2}\)
Mencari Q
\(Q=\frac{\partial f}{\partial y}\)
\(f(x,y)=\int(3x+4y)dy\)
\(3xy+2y^{2}+c(x)\)
\(P=\frac{\partial f}{\partial x}=2x+3y\)
\(3y+{c}'(x)=2x+3y\)
\({c}'=2x\)
\(c(x)=\int2x=x^{2}\)
\(f(x,y)=3xy+2y^{2}+x^{2}\)
Jadi:
\(P=(x,y)=x^{2}+3yx+2y^{2}\)
\(Q=(x,y)=3xy+2y^{2}+x^{2}\)
Contoh 2:
\(2.(x^{2}+2y)dx+(y^{3}+2x)dy=0\)
Jawaban:
\(\frac{\partial M}{\partial y}=2\)
\(\frac{\partial N}{\partial x}=2\)
Karena \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\) maka, persamaan diatas merupakan PD Eksak
\(\mu (x,y)=\int M(x,y)dx+\phi (y)\)
\(=\int x^{2}+2y dx+\phi (y)\)
\(\frac{1}{3}x^{3}+2xy+\phi (y)\)
\(\frac{\partial \mu}{\partial y}=2x+{\phi }'(y)=N(x,y)\)
\(=2x+{\phi }'(y)=y^{3}+2x\)
\(\phi (y)=y^{3}\)
\(\phi (y)=\int y^{3}dy=\frac{1}{4}y^{4}\)
Jadi penyelesaian PD\(\mu(x,y)= C\)
\(\frac{1}{3}x^{3}+2xy+\frac{1}{4}y^{4}=c atau 4x^{3}+24xy+3y^{4}= C\)
Contoh 3:
\(3. (2x+3y-2)dx+(3x-4y+1)dy=0\)
Jawaban:
\(M(x,y)=2x+3y-2\Rightarrow \frac{\partial M}{\partial y}=3\)
\(N(x,y)=3x-4y+1\Rightarrow \frac{\partial N}{\partial x}=3\)
Karena \(\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}=3\), maka disebut PD Eksak
\(\frac{\partial \mu}{\partial x}= M(x,y)2x+3y-2\)
\(\mu (x,y)= \int(2x+3y-2)dx+\rho (y)\)
\(= x^{2}+3xy-2x+\rho(y)\)
\(\frac{\partial \mu}{\partial y}=3x+{\rho}'(y)= 3x-4y+1\)
Didapat \({\rho}'(y)=-4y+1\) dan berati \(\rho(y)=-2y^{2}+y\)
Jadi solusi persamaan diferensial eksak \(2x+3y-2)dx + (3x-4y+1)dy=0\) adalah
\(\mu(x,y)= C\)
\(x^{2}+3xy-2x)+-2y^{2}+y)= C\)
\(x^{2}+3xy-2x-2y^{2}+y= C, C\) konstanta sembarang
Persamaan diferensial eksak
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