Understanding Nonlinear ODE Solutions and Examples
Dive into a comprehensive guide on solving nonlinear ordinary differential equations (ODEs) with detailed examples and solutions. Explore various methods and techniques to tackle ODEs efficiently and enhance your mathematical skills.
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dy dx+ dy dx= a y = f( x) f( x,y) 5
2 d y dx dy dx + + = p Qy f( x) 2 2 d y dx dy dx p( x,y) Q( x,y) y f( x) + + = 2 6
u y dv dt u v 7
2 2 2 u u d v + tv = - = 0 6 1 2 2 y x 2 dt 8
Examples: ( ) dv t dt d x t dt t - ( ) v t = e 2 ( ) ( ) dt dx t - 5 + 2 ( ) x t = cos( ) t 2 9
d v d t M c v cv M = 9.8 - : mass : dragcoefficient :velocity 10
dv c = 8 . 9 v dt M d v c = 8 . 9 v d t M d v c = 8 . 9 v d t M 11
Examples: ( ) dx t dt d x t dt t - ( ) x t = e 2 ( ) ( ) dt dx t - 5 + 2 ( ) x t = cos( ) t 2 3 - 2 ( ) ( ) dt d x t dt dx t 4 + 2 ( ) = 1 x t 2 12
Example: ( ) dx t dt - t Solution Proof: ( ) dx t dt dx t dt ( ) x t = e + ( ) x t = 0 - t = - e ( ) - - t t + ( ) x t = - + = 0 e e 13
x dx dt = - x Solution: ( ) - - - t t t x c e c e c e = = - That is, are all valid solutions - - t t t = = - 2 L x e ,x e , 15
Examples: ( ) dx t dt d x t dt t - ( ) x t = e 2 ( ) ( ) dt dx t 2 - 5 + 2 ( ) = cos( ) t x t t 2 3 - 2 ( ) ( ) dt d x t dt dx t + ( ) x t = 1 2 16
Examples t dx of nonlinear ODE : ( ) = cos( ( )) 1 x t dt 2 ( 2 ) ( ) d x t dx t = 5 ( ) 2 x t dt dt 2 ( 2 ) ( ) d x t dx t + = ( ) 1 x t dt dt 17
= ( ) cos( 2 ) x t t solution a is to the ODE 2 ( 2 ) d x t + = 4 ( ) 0 x t dt = + functions All of the form ( ) cos( 2 ) x t t c (where a is c constant) real solutions. are 18
2 ( 2 ) d x t + = 4 ( ) 0 x t dt = ) 0 ( x a = ) 0 ( x b 19
2 d y dt dy dt + + = 2 2 t 2 5 t t y e 2 20
+ + = 2 t 2 x x x e + + = 2 t 2 x x x e = 5 . 1 = ) 0 ( x , 1 ) 2 ( x = = ) 0 ( x , 1 ) 0 ( x 5 . 2 23
y = ( ) y f t g = = (let 10 ) y g g 2 d y = 10 2 dt t dy = + 10 (1) t c 1 dt = + + 2 5 (2) y t c t c 1 2 24
IVP: y(t y(t & BVP: y(t y(t = = 0 0 = = 0 5 ) ) = = 0 10 = 0 ) = 100 ) 0 10 = = 0 y( ) y( c = = 0 5 c c 2 2 500 10 100 60 ) c c = - + = = 1 1 1 2 5 60 y t t = - + 2 5 5 y(t ) t t = - + 25
x & = = x x,x 1 2 then x & = x 1 2 & 0 mx cx kx + + = 1 m = = 2 2 1 x & = ( kx - - cx ) 2 1 2 0 0 1 0 x ( ) x ( ) 1 IC: 2 - Dim Euler n v v v v v v 0 y( x x) y( x) f( x,y ) x,y( ) D y + D + = 0 30
Assume m=1,c=1, k=1 (for ease of computation) v x x x v - v v x 2 1 = = x f( x,t ) 2 = f ( kx - - cx ) 1 m - x x 1 2 2 1 2 set t = 0.1 v + - 0 99 0 19 - 1 0 0 - 1 0 1 . v v v 0 1 . - - 0 1 0 0 0 0 0 0 0 1 . x( . ) x( . ) f( x( . ), . ) = + = = - 1 0 - v + + - 1 0 1 . 0 1 . ( - 0 99 0 19 . . v v v 0 2 0 1 0 1 0 1 0 1 f( x( . ), . ) 0 1 . x( . ) x( . ) . = + = = - 1 0 1 . ) - v - 0 19 . ( - 0 971 0 27 . . . v v v 0 3 0 2 0 2 0 2 0 1 . 0 1 . x( . ) x( . ) f( x( . ), . ) = + = = - 0 99 . - 0 19 . . ) 31
35 2 d dt q gsin l + = 0 q 2 35
= at = , or ( ) h t = h h t t h 0 0 0 0 41
dy dt= = ( ) y t y ( , ), f t y t t 0 0 0 46
= ( , ) y F x x y + 1 + 1, i i i i = ( , ) y F x x y + 1 + 1, + 1 i i i i 48
