2 matematikos koliokviumas

Dif. lygtis su atskiriamais kintamais: p1(x)q1(y)dx+p2(x)q2(y)dy=0 / *1/q1(y)p2(x)

Apibrezimas:Dif lygtis, I kuria ieina y’+p(x)Y=q(x)...(1) – tiesine dif lygtis. p(x),q(x) – tolydzios x((a,b); a1(x)y’+b1(x)y=c1(x)...(2); a1(x)(0/(1/a1(x)); y’+y b1(x)/a1(x)=c1(x)/a1(x)

b1(x)/a1(x)=p(x); c1(x)/Q1(x)=q(x), gauname y’+p(x)y=q(x). Jei q(x)=0, tai y’+p(x)y=0...(3)- tiesine homogenine

u=u(x) ir v=v(x) – tai x((a,b); y’=u’v+v’u...(5)

(dv/v=(-p(x)dx, v=e(-p(x)dx+c , kai c=0...(8)

2)Lagranzo(konstantu variavimo): Turime tiesine nehom dif lygti y’+p(x)y=q(x)..(1); y’+p(x)y=0..(2);

dy/dx=-p(x)y; y= e-(p(x)dx+c(x); y’= e-(p(x)dx+c(x), c=c(x)

y’+yp(x)=q(x) yn / y-n ;yn y-n+ y1-np(x)=q(x). Pazymim: y1-n =z...(3) Diferencijuojame:

egzistuoja ðu/ðx=ðu/ðy, du=ðu/ðx+ðu/ðy(dy) ...(3), Is (2) ir (3) gauname: ðu/ðx=P(x,y)...(4)

L[cy]=(cy)(n)+ p1(x)(cy)(n-1) +...+ pn(x)cy=c[y(n)+ p1(x)y(n-1) +...+ pn(x)y]=cL[y]=0

y (n) +P1(x) y (n-1) +P2(x) y (n-2) +...+Pn(x) y=f(x)...(1)

Trurime y’’+p1(x)y’+p2(x)y=f(x)...(1) antors eiles tiesine nehomogenine lygtis L[x]=f(x); Sudarome: L[x]=0; y’’+p1(x)y’+p2(x)y=0 isprendziame ir randame bendra sprendini. Y=C1y1+C2y2- bendras spr. (2); y=C1(x)y1(x)+C2(x)y2(x)...(3)

Y’=C’1(x)y1(x)+C1(x)y’1(x)+C’2(x)y2(x)+C2(x)y’2(x); Pagal Lagranza: C’1(x)y1(x)+C’2(x)y2(x)=0 y’= C’1(x)y’1(x)+C’2(x)y’2(x)...(4);