CAS Project 16

(%i1)

kill ( all ) ;

\[\operatorname{ }\ensuremath{\mathrm{done}}\]

Persamaan differensial orde 2 yang akan dicari solusinya menggunakan transformasi Laplace

(%i1)

pdbo2 : a · ' diff ( x ( t ) , t , 2 ) + b · ' diff ( x ( t ) , t , 1 ) + c = %e ^ t ; /*persamaan differensial orde 2; a, b dan c adalah konstanta*/

\[\operatorname{(pdbo2) }a\, \left( \frac{{{d}^{2}}}{d {{t}^{2}}} \operatorname{x}(t)\right) +b\, \left( \frac{d}{d t} \operatorname{x}(t)\right) +c={{\% e}^{t}}\]

Transformasi Laplace pdb tersebut

(%i2)

trlap : laplace ( pdbo2 , t , p ) ; /*transformasi Laplace PDBO2 tersebut, variabel lama adalah t dan variabel baru adalah p*/

\[\operatorname{(trlap) }a\, \left( -\left. \frac{d}{d t} \operatorname{x}(t)\right|_{t=0}+{{p}^{2}} \operatorname{laplace}\left( \operatorname{x}(t)\operatorname{,}t\operatorname{,}p\right) -\operatorname{x}(0) p\right) +b\, \left( p \operatorname{laplace}\left( \operatorname{x}(t)\operatorname{,}t\operatorname{,}p\right) -\operatorname{x}(0)\right) +\frac{c}{p}=\frac{1}{p-1}\]

Masukkan syarat awal, yaitu x(0)=0 dan x'(0)=0

(%i3)

syawal : subst ( [ x ( 0 ) = 0 , at ( ' diff ( x ( t ) , t ) , t = 0 ) = 0 ] , trlap ) ; /*masukkan syarat awal*/

\[\operatorname{(syawal) }a {{p}^{2}} \operatorname{laplace}\left( \operatorname{x}(t)\operatorname{,}t\operatorname{,}p\right) +b p \operatorname{laplace}\left( \operatorname{x}(t)\operatorname{,}t\operatorname{,}p\right) +\frac{c}{p}=\frac{1}{p-1}\]

Cari solusi persamaan tersebut, nyatakan laplace(x(t),t,p) sama dengan sesuatu

(%i4)

solusi : solve ( syawal , ' laplace ( x ( t ) , t , p ) ) ; /*selesaikan persamaan linier yang dihasilkan*/

\[\operatorname{(solusi) }\left[ \operatorname{laplace}\left( \operatorname{x}(t)\operatorname{,}t\operatorname{,}p\right) =-\frac{\left( c-1\right) p-c}{a {{p}^{4}}+\left( b-a\right) {{p}^{3}}-b {{p}^{2}}}\right] \]

Gunakan invers transformasi Laplace

(%i5)

solusi_akhir : ilt ( solusi [ 1 ] , p , t ) ; /*gunakan inverse transformasi laplace untuk mendapatkan solusi x(t)*/

\[\operatorname{(solusi\_ akhir) }\operatorname{x}(t)=-\frac{\left( \left( {{a}^{2}} b+{{a}^{3}}\right) c-{{a}^{2}} b\right) {{\% e}^{-\frac{b t}{a}}}}{a\, \left( {{b}^{3}}+a {{b}^{2}}\right) }+\frac{{{\% e}^{t}}}{b+a}-\frac{c t}{b}+\frac{a c-b}{{{b}^{2}}}\]

Misalkan ingin dihitung jika diketahui nilai a = 1, b = 2 dan c = 3

(%i6)

solkhus : subst ( [ a = 1 , b = 2 , c = 3 ] , rhs ( solusi_akhir ) ) ;

\[\operatorname{(solkhus) }\frac{{{\% e}^{t}}}{3}-\frac{7 {{\% e}^{-2 t}}}{12}-\frac{3 t}{2}+\frac{1}{4}\]

(%i7)

define ( x ( t ) , solkhus ) ;

\[\operatorname{ }\operatorname{x}(t)\operatorname{:=}\frac{{{\% e}^{t}}}{3}-\frac{7 {{\% e}^{-2 t}}}{12}-\frac{3 t}{2}+\frac{1}{4}\]

(%i8)

x ( 1 ) ;

\[\operatorname{ }\frac{\% e}{3}-\frac{7 {{\% e}^{-2}}}{12}-\frac{5}{4}\]

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