Differential Equations : Undetermined Coefficients , Variation Of Parameters And Laplace Transform

DIFFERENTIAL EQUATIONS El presente documento contiene la solución de una ecuación diferencial por medio de variación de parámetros , coeficientes indeterminados y transformada de Laplace. Gracias al ingeniero Ricardo Lopez, profesor de física de la universidad Tecnológica de Pereira por la solución. Cualquier pregunta puede ser realizada a [email protected] ABSTRACT : The following document containt describe how a differential equation can be solved by three methods: Undetermined coefficients , variation of parameters and Laplace Transform. Thanks to proffesor Ricardo Lopez for the solutions, he currently belongs to physics department from UNIVERSIDAD TECNOLOGICA DE PEREIRA . Any question can be done to [email protected] 2 The equation : y ' ' ( x) + 4 y ( x) = cos x

SOLUTION USING VARIATION OF PARAMETERS The solution of the homogeneous equation is : yh = C1 cos(2 x ) + C2 sin( 2 x)

y p = u1 ( x) cos(2 x) + u2 ( x) sin( 2 x ) Wy1, y 2 =

cos(2 x) sin(2 x ) =2, − 2 sin(2 x ) 2 cos(2 x)

du1 sin( 2 x) du cos(2 x) =− cos 2 x ,      2 = cos 2 x dx 2 dx 2

u1 ( x) = −

1 1 1  1 + cos(2 x )  sin(2 x) dx = cos(2 x) + 1 cos(4 x ) ∫ 2 2 8 32  

cos(2 x ) cos( x)dx 1 1 1  1 + cos(2 x)  2 = ∫ cos(2 x) dx = ∫ cos(2 x )dx + ∫ cos (2 x) dx = 2 2 2 4 4   1 1 x 1 1 sin( 2 x) + ∫ (1 + cos(4 x)dx = + sin( 2 x ) + 8 8 8 8 32

u2 ( x) = ∫

y p = u1 ( x ) cos(2 x) + u2 ( x) sin( 2 x)

replacing

1 1 1 1 1 ∴ y p = cos 2 (2 x) + cos(2 x) cos(4 x) + sin 2 (2 x ) + x sin( 2 x) + sin( 4 x) sin( 2 x) 8 32 8 8 32 Then grouping the terms :

yp =

1 1 1 + x sin( 2 x) + cos(2 x) 8 8 32

The last one it’s exclude because it’s already in the homogeneous solution.

SOLUTION USING UNDETERMINED COEFFICIENTS y ' ' ( x) + 4 y ( x) = cos 2 x =

1 (1 + cos( 2 x ) ) 2

Auxiliary equation: r 2 + 4 = 0

r = ±2i

yh = C1 cos(2 x ) + C2 sin( 2 x) 1 (1 + cos(2 x) ) = 0 2 D( D 2 + 4)( D 2 + 4) y = 0            ,  D( D 2 + 4) 2 y = 0 D( D 2 + 4)

Auxiliary equation: r ( r 2 + 4) 2 = 0

r = 0  ,  r = ±2i

Particulary equation structure:

y p ( x) = A + Bx cos(2 x ) + Cx sin( 2 x ) y ' p ( x) = B cos(2 x) − 2 Bx sin( 2 x) + C sin( 2 x ) + 2Cx cos(2 x) y ' ' p ( x) = −3B sin( 2 x) + 4C cos(2 x ) − 4 Bx cos(2 x) − 4Cx sin(2 x) Replacing into the differential equation :

− 3B sin(2 x) + 4C cos(2 x ) − 4 B cos(2 x) − 4Cx sin(2 x) + 4 A + 4 Bx cos(2 x) + 4Cx sin(2 x ) = 1 1 1 1 ∴  A =   ,  B = 0   ,  C =     y p ( x) = + x sin(2 x) 8 8 8 8

General solution : y ( x) =C 1cos(2 x) + C2 sin(2 x ) +

1 1 + x sin(2 x) 8 8

1 1 + cos(2 x) 2 2

SOLUTION USING LAPLACE TRANSFORM x + 6 x + 18 x = cos 2t , x(0) = 1  ,  x (0) = −1 m 2 + 6m + 18 = 0    m = −3 ± 3i xh (t ) = −3t (C1 cos(3t ) + C2 sin(3t )) x p (t ) = A cos(3t ) + B sin(3t ) s 2 x ( s ) − s + 1 + 6 sx ( s ) − 6 + 18 x( s ) =

{

}

s s +4 2

s s +4 s As + B Cs + D = 2 + 2 2 2 ( s + 6s + 18)( s + 4) s + 4 s + 6 s + 18

x( s) s 2 + 6s + 18 = s + 5 +

2

( A + C ) S 3 + ( B + 6 A + D) S 2 + (6 B + 18 A + 4C ) S + 18 B + 4 D = 0 9 A + C = 0     ∴  C = − A  ,  18 B + 4 D = 0    D = − B   2 9 7 B + 6 A + D = B + 6 A − B = 0   ∴  − B + 6 A = 0 2 2 14 A + 6 B = 1 7 9  12  54 7     B = −    = −    C = −    170 2  170  170 170 2 7 6 7 11 − 3t x(t ) = − 3t cos(3t ) + − 3t sin(3t ) + cos(2t ) + sin( 2t ) − cos(3t )− 3t −  sin(3t ) 3 170 170 170 170 307 163 − 3t 7 6 x(t ) =  cos(3t ) + 3 − 3t sin(3t ) + cos(2t ) + sin(2t ) 170 170 170 170 A=

SOLUTION USING THE TRADITIONAL WAY

x + 6 x + 18 x = cos(3t )        x(0) = 1  ,  x (0) = −1 m 2 + 6m + 18 = 0   ,   m = −3 ± 3i xh (t ) = − 3t (C1 cos(3t ) + C2 sin(3t )) x p (t ) = A cos(2t ) + B sin( 2t ) x p (t ) = −2 A sin( 2t ) + 2 B cos(2t ) x p (t ) = −4 A cos(2t ) − 4 B sin( 2t ) (−4 A + 12 B + 18 A) cos(2t ) + (−4 A − 22 A + 18 B ) sin( 2t ) = cos(2t ) ∴    14 A + 12 B = 1 − 12 A + 14 B = 0 1 12 A=

14

1

− 12 0 12 14 0 14 =     ,    B = = 14 12 340 340 340 − 12 14

x(t ) = − 3t (C1 cos(3t ) + C2 sin(3t )) +

14 12 cos(2t ) + sin(2t ) 340 340

x (t ) = −3− 3t (C1 cos(3t ) + C2 sin(3t )) + − 3t (−3C1 sin(3t ) + 3C2 cos(3t )) − 1 = C1 +

14 7 163 = C1 +   ,  C1 = 340 170 170

307 12 12 489 307 − 1 = −3C1 + 3C2 +   ,  3C2 = −1 − + =   ∴  C2 = 13 170 170 170 170 170 307    163  7 6 − 3t 3  x(t ) = 3 cos(3t ) + sin(3t )  + cos(2t ) + sin(2t ) 170 170  170  170    

28 24 sin(2t ) + cos(2t ) 340 340