> restart; > with(linalg): Warning, the protected names norm and trace have been redefined and unprotected > with(inttrans): Warning, the name hilbert has been redefined > with(student): ELSORENDU ALLANDO EGYUTTHETOS LIN. DIFF. EGYENLET REND- SZER y =y2+y3+x, y2 =y-y3+exp(2, y3 =y+y2-x y(0)=y2(0)=y3(0)=/2 DIREKT MEGOLDAS Matrixos alakban Y=A.y+b > Y:=matrix([[diff(y(,],[diff(y2(,],[diff(y3(,]]); x y( Y := x y2( x y3( > A:=matrix([[0,,],[,0,-],[,,0]]); A := 0 0 0 > y:=matrix([[y(],[y2(],[y3(]]); y := y( y2( y3(
> b:=matrix([[x],[exp(2*],[-x]]); x b := (2 e x > eh:=evalm(y=multiply(a,y)); x y( y2( + y3( eh := x y2( = y( y3( x y3( y( + y2( > ei:=evalm(y=multiply(a,y)+b); ei := x y( x y2( x y3( = y2( + y3( + x y( y3( + e y( + y2( x (2 Homogen megoldasa: > s:=eigenvects(a); s := [,, {[,, 0]}], [,, {[, 0, ]}], [0,, {[,, ]}] Az elso elem a sajatertek, a masodik a multiplicitasa, a harmadik a sajatvektor., tehat > s:=s[][]; > s2:=s[2][]; > s3:=s[3][]; > v:=s[][3][]; > v2:=s[2][3][]; > v3:=s[3][3][]; s := s2 := s3 := 0 v := [,, 0] v2 := [, 0, ] v3 := [,, ] > V:=convert(v,matri;
V := > V2:=convert(v2,matri; > V3:=convert(v3,matri; V2 := 0 V3 := 0 > e:=evalm(exp(s**v); e := e( e ( 0 > e2:=evalm(exp(s2**v2); e2 := > e3:=evalm(exp(s3**v3); Tehat az alaprendszer: e3 := ex 0 e x > F:=evalm(augment(e,e2,e3)); e ( e x F := e ( 0 0 e x A homogen altalanos: y ha = F.c > c:=matrix([[c],[c2],[c3]]); c := c c2 c3 > Z:=multiply(F,c);
azaz Z := e( c + e x c2 + c3 e ( c c3 e x c2 + c3 > Y:=Z[,]; > Y2:=Z[2,]; > Y3:=Z[3,]; Y := e ( c + e x c2 + c3 Y2 := e ( c c3 Y3 := e x c2 + c3 Ellenorzes: behelyettesitjuk a homogen egyenletbe: > evalm(subs(y(=y,y2(=y2,y3(=y3,eh)); x ( e( c + e x c2 + c3 ) = x (e( c c3 ) x (ex c2 + c3 ) e ( c + e x c2 e ( c e x c2 Inhomogen, allandok varialasa: y ia = F. d(, ahol d( ugy kaphato, hogy megoldandjuk az F.d (=b egyenletrendszert. Az egyenletrendszer matrixa: > M:=evalm(augment(F,b)); e ( e x x M := e ( 0 e 0 e x x (2 Gauss eliminacio: > MM:=simplify(expand(gaussjord(M))); 0 0 2 x e x MM := 0 0 (x + e (2 ) e ( 0 0 (2 2 x e
Innen leolvashatoak a d(, d2(, d3( derivaltjai, tehat megkaphatoak maguk a d, d2, d3 : > d:=simplify(expand(subs(t=x,int(mm[,4],x=0..t)))); d := 2 x e x + 2 e x 2 > d2:=simplify(expand(subs(t=x,int(mm[2,4],x=0..t)))); d2 := e ( x e ( + e x > d3:=simplify(expand(subs(t=x,int(mm[3,4],x=0..t)))); d3 := x 2 2 + 2 Vektort csinalunk a d-bol, hogy az alaprendszer matrixaval osszeszorozhassuk (ez csak Maple ugyeskedes): > d:=convert([d,d2,d3],vector); [ d := 2 x e x + 2 e x 2, e ( x e ( + e x, x 2 2 + ] 2 > dd:=simplify(evalm(augment(d))); dd := 2 x e x + 2 e x 2 e ( x e ( + e x x 2 2 + 2 Es ime a partikularis megoldas vektora: > S:=simplify(expand(multiply(F,dd))); x 5 2 + 2 e( + 2 x 2 S := 2 x + 3 2 2 e( + x 2 + 2 2 x + 2 x 2 Es ezzel az inhomogen partikularis > w:=s[,];
w := x 5 2 + 2 e( + 2 x 2 > w2:=s[2,]; w2 := 2 x + 3 2 2 e( + x 2 + 2 > w3:=s[3,]; w3 := 2 x + 2 x 2 Ellenorzes: behelyettesitjuk az inhomogen egyenletbe: > evalm(subs(y(=w,y2(=w2,y3(=w3,ei)); x (x 5 2 + 2 e( + 2 x 2 ) 2 x + 2 e ( (2 + e x ( 2 x + 3 2 2 e( + x 2 + 2 ) x ( 2 x + 2 x 2 ) = 2 x 2 + 2 e ( + e + e (2 2 x (2 Tehat az altalanos megoldas: > z:=y+w; z := e ( c + e x c2 + c3 + x 5 2 + 2 e( + 2 x 2 > z2:=y2+w2; z2 := e ( c c3 2 x + 3 2 2 e( + x 2 + 2 > z3:=y3+w3; z3 := e x c2 + c3 2 x + 2 x 2 Az altalanos megoldas ellenorzese: > evalm(subs(y(=y+w,y2(=y2+w2,y3(=y3+w3,ei));
x ( e( c + e x c2 + c3 + x 5 2 + 2 e( + 2 x 2 ) x (e( c c3 2 x + 3 2 2 e( + x 2 + 2 ) x (ex c2 + c3 2 x + 2 x 2 ) e ( c 2 x + 2 e ( + e (2 + e x c2 e ( c + 2 x 2 + 2 e ( (2 + e e x c2 + e (2 2 x = Kezdet ertek: z(0)=z2(0)=z3(0)=/2 : > simplify(subs(x=0,z))=/2; c + c2 + c3 = 2 > simplify(subs(x=0,z2))=/2; c c3 = 2 > simplify(subs(x=0,z3))=/2; c2 + c3 = 2 A megoldando linearis egyenletrendszer tehat: > K:=matrix([[-,,,/2],[,-,0,/2],[0,,,/2]]); 2 K := 0 2 0 2 > KV:=gaussjord(K);
0 0 0 KV := 0 0 2 0 0 Ezzel megvannak az alkalmas konstansok, igy a kezdeti ertek problema megoldasa : > W:=subs(c=0,c2=-/2,c3=,z); W := 3 2 2 ex + x + 2 e ( + 2 x 2 > W2:=subs(c=0,c2=-/2,c3=,z2); W2 := 2 2 x 2 e( + x 2 + 2 > W3:=subs(c=0,c2=-/2,c3=,z3); W3 := 2 ex + 2 x + 2 x 2 > restart; > with(linalg): Warning, the protected names norm and trace have been redefined and unprotected > with(inttrans): Warning, the name hilbert has been redefined > with(student): MEGOLDAS LAPLACE-val > de:=diff(y(,=y2(+y3(+x; de := x y( = y2( + y3( + x > de2:=diff(y2(,=y(-y3(+exp(2*; de2 := x y2( = y( y3( + > de3:=diff(y3(,=y(+y2(-x; de3 := x y3( = y( + y2( x
Ezek Laplace transzformaltjai: > alias(u=laplace(y(,x,s)); Point, U > alias(v=laplace(y2(,x,s)); Point, U, V > alias(w=laplace(y3(,x,s)); Point, U, V, W > alias(a=y(0)); > alias(b=y2(0)); > alias(c=y3(0)); Point, U, V, W, a Point, U, V, W, a, b Point, U, V, W, a, b, c > L:=laplace(de,x,s); > L2:=laplace(de2,x,s); L := s U a = V + W + s 2 L2 := s V b = U W + s 2 > L3:=laplace(de3,x,s); L3 := s W c = U + V s 2 Most a Laplace transzformaltakbol allo linearis egyenletrendszert kell megoldani: > G:=solve({L,L2,L3},{U,V,W});
G := {V = b s4 + s 3 c s 3 b s 3 + a s 3 2 b s 2 + 2 c s 2 2 a s 2 + s 2 + 2 s 4 (s + ) (s 2) s 3, W = a s3 3 c s 3 + 4 s 2 a s 2 + 2 c s 2 4 + c s 4 + b s 3 2 b s 2 s 3 ( 3 s + 2 + s 2, U = ( 4 3 c s 3 ) + a s 3 2 b s 2 b s 3 + c s 4 + b s 4 2 a s 4 + a s 5 + 4 s + 2 s 3 2 s 2 + 2 c s 2 2 a s 2 )/(s 3 ( 3 s + 2 + s 2 ) (s + ))} Egyenletrendszer megoldas ellenorzes: > A:=matrix([[s,-,-,a+/s^2],[-,s,,b+/(s-2)],[-,-,s,c-/s^2]]): > gaussjord(a): Parcialis tortekre bontva (ez persze itt nem kell, csak ha kezzel szamoljuk a visszatranszformaltat): > convert(g[],parfrac,s); V = 2 + c a (2 a + 3 + 2 b 2 c) + 2 s + s 2 + 2 2 s s 2 + 2 s 3 > convert(g[2],parfrac,s); W = b + a + s + 2 s 2 2 > convert(g[3],parfrac,s); U = 2 + c a s + 2 5 2 c + 2 a + 2 b s 2 c + 2 b + 2 a s s 2 2 s 3 + s 2 2 s 3 + b + a + s + 2 s 2 Ezeket visszatranszformalva: > z:=invlaplace(g[],s,; z := y2( = ( 2 + c a) e ( + 2 + a + 3 2 + b c 2 x + x2 > z2:=invlaplace(g[2],s,;
z2 := y3( = b a + c 2 x x2 + (b + a) e x + 2 > z3:=invlaplace(g[3],s,; z3 := y( = 5 2 + c a b + x x2 + (b + a) e x + 2 + (a + 2 c) e ( Ahhoz, hogy a szokasos jelolessel (amit a kozvetlen megoldasnal hasznaltunk) kapjuk a megoldast, az egyutthatokat alkalmasan atjeloljuk: > e:=c_=-5/2-b+c-a+3; e := c = 2 b + c a > e2:=c_2=b+a; e2 := c 2 = b + a > e3:=c_3=2-c+a; e3 := c 3 = a + 2 c > s:=solve({e,e2,e3},{a,b,c}); s := {a = c 2 + c 3 5 2 + c, b = c 3 + 5 2 c, c = c 2 + c 2 } > s[]; > s[2]; > s[3]; a = c 2 + c 3 5 2 + c b = c 3 + 5 2 c c = c 2 + c 2 Ezeket visszahelyettesitve a megoldasba megkapjuk az altalanos megoldast (valamiert nem okkvetlenul sorremdben adja meg): > m:=subs(a=c_2+c_3-5/2+c_,b=-c_3+5/2-c_,c=c_-/2+c_2,z); m := y2( = c 3 e ( + 2 + 2 c 2 x + x 2
> m2:=subs(a=c_2+c_3-5/2+c_,b=-c_3+5/2-c_,c=c_-/2+c_2,z2); m2 := y3( = + c x x 2 + c 2 e x + 2 > m3:=subs(a=c_2+c_3-5/2+c_,b=-c_3+5/2-c_,c=c_-/2+c_2,z3); m3 := y( = c 3 + x x 2 + c 2 e x + 2 + c 3 e ( (b) Kezdeti ertek problema: y(0)=y2(0)=y3(0)=/2 A kezdeti ertek problema megoldasat termeszetesen a z, z2, z3-bol az a=b=c=/2 helyettesitesekbol kovetlenul megkaphatjuk: > k:=subs(a=/2, b=/2, c=/2, z); k := y2( = 2 e ( + 2 + 2 2 x + x 2 > k2:=subs(a=/2, b=/2, c=/2, z2); k2 := y3( = x x 2 + e x + 2 > k3:=subs(a=/2, b=/2, c=/2, z3); k3 := y( = 3 + x x 2 + e x + 2 + 2 e ( De persze kozvetlenul az elejen is elvegezhettuk volna a helyettesitest: > restart; > with(linalg): Warning, the protected names norm and trace have been redefined and unprotected > with(inttrans): Warning, the name hilbert has been redefined > with(student): > de:=diff(y(,=y2(+y3(+x; de := x y( = y2( + y3( + x > de2:=diff(y2(,=y(-y3(+exp(2*; de2 := x y2( = y( y3( + > de3:=diff(y3(,=y(+y2(-x;
de3 := x y3( = y( + y2( x Ezek Laplace transzformaltjai: > alias(u=laplace(y(,x,s)); Point, U > alias(v=laplace(y2(,x,s)); Point, U, V > alias(w=laplace(y3(,x,s)); Point, U, V, W > alias(a=y(0)); > alias(b=y2(0)); > alias(c=y3(0)); Point, U, V, W, a Point, U, V, W, a, b Point, U, V, W, a, b, c > L:=laplace(de,x,s); > L2:=laplace(de2,x,s); L := s U a = V + W + s 2 L2 := s V b = U W + s 2 > L3:=laplace(de3,x,s); L3 := s W c = U + V s 2 > k:=subs(a=/2, b=/2, c=/2, L); k := s U 2 = V + W + s 2 > k2:=subs(a=/2, b=/2, c=/2, L2); k2 := s V 2 = U W + s 2 > k3:=subs(a=/2, b=/2, c=/2, L3); k3 := s W 2 = U + V s 2 Most a Laplace transzformaltakbol allo linearis egyenletrendszert
kell megoldani: > G:=solve({k,k2,k3},{U,V,W}); G := {U = 2 W = 2 s 5 + s 3 6 s 2 + 8 s 8 ( + s) (s 2) (s ) s 3, V = s 4 + s 3 + 4 s 8 2 s 3 (s 2) ( + s), 8 s 3 + 8 s 2 s 2 + s 4 s 3 (2 3 s + s 2 } ) Egyenletrendszer megoldas ellenorzes: > A:=matrix([[s,-,-,a+/s^2],[-,s,,b+/(s-2)],[-,-,s,c-/s^2]]): > gaussjord(a): Parcialis tortekre bontva (ez persze itt nem kell, csak ha kezzel szamoljuk a visszatranszformaltat): > convert(g[],parfrac,s); U = 2 + s + 2 s 2 + s 2 s 3 + s 2 3 s > convert(g[2],parfrac,s); V = 2 + s + 2 s 2 + 2 s 3 2 s 2 + 2 s > convert(g[3],parfrac,s); W = 2 s 2 + s 2 s 3 s 2 s Ezeket visszatranszformalva: > z:=invlaplace(g[],s,; z := y( = 2 e ( + 2 + e x x 2 + x 3 > z2:=invlaplace(g[2],s,;
z2 := y2( = x 2 2 x + 2 + 2 2 e ( > z3:=invlaplace(g[3],s,; z3 := y3( = x 2 x + e x + 2 ELLENORZES Altalanos megoldas: > ds:=dsolve({de,de2,de3},{y(,y2(,y3(}); ds := {y3( = e x C2 + 2 x x 2 + C3 + 2, y2( = e ( C 2 x + 2 + x 2 C3, y( = e x C2 e ( C x 2 + 2 + x + C3 } Tehat: > ds[]; y3( = e x C2 + 2 x x 2 + C3 + 2 > ds[2]; y2( = e ( C 2 x + 2 + x 2 C3 > ds[3]; y( = e x C2 e ( C x 2 + 2 + x + C3 Az osszehasonlitashoz (lehet, hogy maskor mas sorrendben sorolja fel, ugyhogy az indexek valtozhatnak, meg kell nezni, hogy melyik komponens melyik megoldast adja). A megfelelo c-k megvalasztasaval lathatoan) azonos alakuak.
Es a kezdeti ertek problema megoldasanak ellenorzese: > dsk:=dsolve({de,de2,de3,y(0)=/2,y2(0)=/2,y3(0)=/2}, > {y(,y2(,y3(}); dsk := {y( = 2 e ( + 2 + e x x 2 + x 3, y2( = x 2 2 x + 2 + 2 2 e (, y3( = x 2 x + e x + 2 } > dsk[]; y( = 2 e ( + 2 + e x x 2 + x 3 > dsk[2]; y2( = x 2 2 x + 2 + 2 2 e ( > dsk[3]; y3( = x 2 x + e x + 2 Es az osszehasonlitas (ez konnyebb): > z; y( = 2 e ( + 2 + e x x 2 + x 3 > z2; y2( = x 2 2 x + 2 + 2 2 e ( > z3; y3( = x 2 x + e x + 2