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Persamaan
(1+e^x)\frac{dy}{dx}+e^x y=0
y(0)=1

Kita susun ulang persamaanya
(1+e^x)\frac{dy}{dx}=-e^x y
(1+e^x) dy=-e^x y dx
\frac{1}{y} dy=-\frac{e^x}{1+e^x} dx
\int \frac{1}{y} dy=-\int \frac{e^x}{1+e^x} dx
Ln(y)+c_1=-Ln(1+e^x)+c_2
Ln(y)=-Ln(1+e^x)+Ln(c)
Ln(y)=Ln(\frac{c}{1+e^x})
y=\frac{c}{1+e^x}

Sekarang kita masukkan y(0)=1
1=\frac{c}{1+e^0}
1=\frac{c}{2}
c=2
Maka
y=\frac{2}{1+e^x}

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