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

3. 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. 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. 5 Effect of Changing  t =  ; y(  )=36.8

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

7. 7 Particular Solution Example:

8. 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. 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. 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. 11 This equation will be called exact if and nonexact otherwise P.Diff. Exact (E) mempunyai penyelesaian Umum :

12. 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. 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. 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. 15 Contoh Solution: which clearly implies that the equation is not exact.

16. 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