1 Penyelesaian dari Persamaan differensial order satu For 1 st order systems, this general form can be rewritten as follows: The constant (  ) is known.

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1 Penyelesaian dari Persamaan differensial order satu For 1 st order systems, this general form can be rewritten as follows: The constant (  ) is known as the time constant of the 1 st order system.

2 Solving 1 st Order Systems The general solution for a homogeneous 1 st order system:

3 Solving 1 st Order Systems The value of the constant (A) can be determined by the initial condition or boundary condition. (Note that there is no particular solution for a homogeneous system.)

4 Solving 1 st Order Systems The time constant (  ) measures how fast the 1 st order system responds. Specifically, at t= , the response, y(t), will have moved ~63.2% of the way between its initial value at t=0 and its final value at t= .

5 Effect of Changing  t =  ; y(  )=36.8

6 Solving 1 st Order Systems A particular solution for a non-homogeneous 1 st order system can be found as follows:

7 Particular Solution Example:

8 Solving 1st Order Systems So, to solve a 1 st order differential equation: 1) find the particular solution (if the problem is non-homogeneous) and solve for all constants 2) find the general solution (this will be of the same form for all 1 st order differential equations) 3) add the particular solution to the general solution to get the total solution 4) solve for the remaining constant using the initial condition or boundary condition.

9 Homogeneous Equations is homogeneous if the function f(x,y) is homogeneous, that is- f (tx, ty ) = f( x, y ) for any number t To solve this type of equation we make use of a substitution. Indeed, consider the substitution atau

10 Contoh. 1. Selesaikan P.D : ( 2x – y ) dy = ( x – 2y ) dx ruas kiri dan kanan masing-masing dibagi dgn x substitusi y = vx dan dy = x dv + v dx ( 2 – v ) ( x dv + v dx ) = ( 1 – 2 v ) dx x ( 2 – v ) dv = ( ( 1 - 2 v ) - v ( 2 – v ) ) dx x ( 2 – v ) dv = ( 1 - 2 v - 2v + v 2 ) dx - ½ ln ( 1 – 4v + v 2 ) = ln x + ln c 1 x 2 ( 1 - 4y/x + y 2 /x 2 ) = c atau x 2 - 4 xy + y 2 = c

11 This equation will be called exact if and nonexact otherwise P.Diff. Exact (E) mempunyai penyelesaian Umum :

12 Contoh soal. 1. Selesaikan P.D : ( x + y – 10 ) dx + ( x – y – 2 ) dy = 0 Jawab. M (x, y ) = ( x + y – 10 ) N ( x, y ) = ( x – y – 2 ) jadi exact., sehingga penyelesaian umumnya adalah : ½ x 2 + xy – 10 x - ½ y 2 – 2y = c

13 Integrating Factor Technique Assume that the equation M(x,y) dx + N(x,y)dy = 0 is not exact, that is- Case 1: if the expression is a function of x only, that is, the variable y disappears from the expression

14 In this case, factor integration u is given by Case 2 : Case 2 : If the expression is a function of y only, that is, the variable x disappears from the expression. In this case, the function u is given by

15 Contoh Solution: which clearly implies that the equation is not exact.

16 (Cont. ) Let us find an integrating factor. We have Therefore, an integrating factor u(x) is given by The new equation is The solutions are given by the implicit equation

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