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http://fmipa.uny.ac.id TAKWA, MANDIRI, CENDEKIA PERSAMAAN DIFERENSIAL Pertemuan 9, PD Bernouli Nikenasih Binatari [email protected] http://uny.ac.id TAKWA, MANDIRI, CENDEKIA

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http://fmipa.uny.ac.idTAKWA, MANDIRI, CENDEKIA

PERSAMAAN DIFERENSIALPertemuan 9, PD Bernouli

Nikenasih [email protected]

http://uny.ac.idTAKWA, MANDIRI, CENDEKIA

http://fmipa.uny.ac.idTAKWA, MANDIRI, CENDEKIA

1β€’ Review PD Homogen

2β€’ Review PD Linear

3β€’ PD Bernoulli

PRESENTATION PARTS

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Review PD Homogen

Definisi homogen

A function 𝑓 of two variables x and y is saidhomogeneous if it satisfies

𝑓 𝑑π‘₯, 𝑑𝑦 = 𝑓 π‘₯, 𝑦for any number t.

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Definition

A function M of two variables x and y is said to behomogenous of degree-n if it satisfies

𝑀 𝑑π‘₯, 𝑑𝑦 = 𝑑𝑛𝑀 π‘₯, 𝑦

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Definition

A first differential equation

𝑀 π‘₯, 𝑦 𝑑π‘₯ + 𝑁 π‘₯, 𝑦 𝑑𝑦 = 0 or𝑑𝑦

𝑑π‘₯= 𝑓(π‘₯, 𝑦)

is called homogeneous differential equation if Mand N are homogeneous functions of the samedegree n or f is a homogenous function.

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Menyelesaikan PD Homogen

PD Homogen (dalam x dan y)

Transform y = vx

PD separabel (dalam v dan x)

Selesaikan PD separabel

Inverse transform v = y/x

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PD Linear

Definition

A first-order ordinary differential equation is linearin the dependent variable y and the independentvariable x if it is, or can be, written in the form

𝑑𝑦

𝑑π‘₯+ 𝑃 π‘₯ 𝑦 = 𝑄 π‘₯

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Menyelesaikan PD Linear

Suppose the integrating factor

πœ‡ π‘₯ = 𝑒 𝑃 π‘₯ 𝑑π‘₯

The general solution of linear differential equationas follows

𝑦 = πœ‡βˆ’1 π‘₯ πœ‡ π‘₯ 𝑄 π‘₯ 𝑑π‘₯ + 𝑐

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PD Bernoulli

Definition

An equation of the form𝑑𝑦

𝑑π‘₯+ 𝑃 π‘₯ 𝑦 = 𝑄 π‘₯ 𝑦𝑛

is called a Bernoulli differential equation.

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TheoremSuppose n β‰  0 or n β‰  1, then the transformation 𝑣 =𝑦1βˆ’π‘› reduces the Bernoulli equation

𝑑𝑦

𝑑π‘₯+ 𝑃 π‘₯ 𝑦 = 𝑄 π‘₯ 𝑦𝑛

into a linear equation in v.

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ExampleSolve the differential equation below

π‘₯𝑑𝑦

𝑑π‘₯+ 𝑦 = π‘₯2𝑦2

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Answer :To solve the differential equations above, first wehave to divided both side with π‘₯𝑦2. Here we get,

π‘¦βˆ’2𝑑𝑦

𝑑π‘₯+ π‘₯βˆ’1π‘¦βˆ’1 = π‘₯

Suppose 𝑣 = π‘¦βˆ’1 then𝑑𝑣

𝑑𝑦= βˆ’π‘¦βˆ’2 or else

𝑑𝑣

𝑑π‘₯=

βˆ’ π‘¦βˆ’2𝑑𝑦

𝑑π‘₯. By here, we get

𝑑𝑣

𝑑π‘₯+ π‘₯βˆ’1𝑣 = π‘₯

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To solve the equation, we use integrating factor

πœ‡ π‘₯ = 𝑒 1π‘₯𝑑π‘₯ = 𝑒ln π‘₯ = π‘₯

Then the solution is

𝑣 = πœ‡βˆ’1 π‘₯ πœ‡ π‘₯ 𝑄 π‘₯ 𝑑π‘₯ + 𝑐

𝑣 = π‘₯βˆ’1 π‘₯2 𝑑π‘₯ + 𝑐 β†’ 𝑣 = π‘₯βˆ’11

3π‘₯3 + 𝑐

=1

3π‘₯2 + 𝑐π‘₯βˆ’1

π‘¦βˆ’1 =1

3π‘₯2 + 𝑐π‘₯βˆ’1

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Solve the following differential equations.

1.𝑑𝑦

𝑑π‘₯βˆ’1

π‘₯𝑦 = π‘₯𝑦2

2.𝑑𝑦

𝑑π‘₯+𝑦

π‘₯= 𝑦2

3.𝑑𝑦

𝑑π‘₯+1

3𝑦 = 𝑒π‘₯𝑦4

4. π‘₯𝑑𝑦

𝑑π‘₯+ 𝑦 = π‘₯𝑦3

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Referensi :

1. Ross, L. Differential Equations. John Wiley &Sons.

http://fmipa.uny.ac.idTAKWA, MANDIRI, CENDEKIA

THANK YOU

Nikenasih [email protected]

Karangmalang Sleman Yogyakarta