Prosím o pomoc s výpočtem
Odpovědi
Diskuze
a) y1=x
y=x
y'=1
y"=0
y" - 2/x * y' + 2/x² * y = 0
0 - 2/x * 1 + 2/x² * x = 0
0 - 2/x + 2/x = 0
================= L=P
b) y2=x²
y=x²
y'=2x
y"=2
y" - 2/x * y' + 2/x² * y = 0
2 - 2/x * 2x + 2/x² * x² = 0
2 - 4 + 2 = 0
============= L=P
B) obecné řešení NLDR:
y" - 2/x * y' + 2/x² * y = 0
substituce: y=z*x
y=z*x
y'=z'*x + z
y"=z"*x + 2z'
y" - 2/x * y' + 2/x² * y = 0
z"*x = 0
z" = 0
z = C1 + C2*x
y=z*x
y1=x
y2=x²
Z výše uvedeného ad a) plyne, je y1 a y2 jsou funkce, které jsou fundamentálním systémem řešení, což bylo v ad b) dokázáno.
y0 = C1*y1 + C2*y2
y0 = C1*x + C2*x²
=================
zkouška:
y = C1*x + C2*x²
y'= C1 + 2*C2*x
y"= 2*C2
y" - 2/x * y' + 2/x² * y = 0
2*C2 - 2/x * (C1 + 2*C2*x) + 2/x² * (C1*x + C2*x²) = 0
L=P
---
C) obecné řešení nehomogenní DR (rovnice s pravou stranou)
použijeme metodu variace konstant:
y = C1(x)*x + C2(x)*x²
metoda variace konstant:
y = C1(x)*x + C2(x)*x²
y'= C1(x) + C1'(x)*x + C2'(x)*x² + 2*C2(x)*x
C1'(x)*x + C2'(x)*x² = 0
y'= C1(x) + 2*C2(x)*x
y"= C1'(x) + 2*C2'(x)*x + 2*C2(x)
y" - 2/x * y' + 2/x² * y = x*e(x)
C1'(x) + 2*C2'(x)*x + 2*C2(x) - 2/x * (C1(x) + 2*C2(x)*x) + 2/x² * (C1(x)*x + C2(x)*x²) = x*e(x)
C1'(x) + 2*C2'(x)*x = x*e(x)
C1'(x)*x + C2'(x)*x² = 0
C1'(x) + 2*C2'(x)*x = x*e(x)
C1'(x) + C2'(x)*x = 0
C1'(x) + 2*C2'(x)*x = x*e(x)
C2'(x)*x = x*e(x)
C2'(x) = e(x)
C2(x) = e(x) + k2
2C1'(x) + 2C2'(x)*x = 0
C1'(x) + 2*C2'(x)*x = x*e(x)
C1'(x) = -x*e(x)
C1(x) = -e(x)*(x-1) + k1
k1=k2=0
Y = C1(x)*x + C2(x)*x²
Y = -e(x)*(x-1)*x + e(x)*x²
Y = x*e(x)
y=y0+Y
=========================
y = C1*x + C2*x² + x*e(x)
=========================
W(x) =((x , x²) , (1, 2x)) = x²
W1(x)=((0 , x²) , ( x*e(x), 2x)) = -x³*e(x)
W2(x)=((x , 0) , (1, x*e(x))) = x²*e(x)
http://www.wolframalpha.com/input/?i=((x+,+x%C2%B2)+,+(1,+2x))%3D
http://www.wolframalpha.com/input/?i=((0+,+x%C2%B2)+,+(+x*e(x),+2x))%3D
http://www.wolframalpha.com/input/?i=((x+,+0)+,+(1,+x*e(x)))%3D
C1'(x)=W1(x)/W(x) = -x³*e(x)/x² = -x*e(x)
C2'(x)=W2(x)/W(x) = x²*e(x)/x² = e(x)
∫x*e(x) = x*e(x)-∫e(x) = x*e(x)-e(x) + C
u'=e(x); u=e(x)
v=x;v'=1
C1(x)=e(x)-x*e(x) = e(x)*(1-x) + k1
C2(x)=e(x) + k2
k1=k2=0
C1(x)=e(x)*(1-x)
C2(x)=e(x)
Y=C1(x)*y1(x) + C2(x)*y2(x)
Y=e(x)*(1-x)*x + e(x)*x²
Y=x*e(x)
y = C1*x + C2*x²
y = y0 + Y
=========================
y = C1*x + C2*x² + x*e(x)
=========================
zkouška:
y = C1*x + C2*x² + x*e(x)
y'= C1 + 2*C2*x + e(x) + x*e(x)
y"= 2*C2 + e(x) + e(x) + x*e(x)
y" - 2/x * y' + 2/x² * y = x*e(x)
2*C2 + e(x) + e(x) + x*e(x)
- 2/x * (C1 + 2*C2*x + e(x) + x*e(x))
+ 2/x² * (C1*x + C2*x² + x*e(x))
= x*e(x)
L=P