[MD-sorular] Size uygun üyeler

[Ege Azuz]

'Homojen olmayan denklemin genel çözümü hom. olanın gen. çöz. ile hom.
olmayanın bir özel çözümünün toplamı olduğundan (aslında çözümler kümesi bir
vektör uzayıdır)'

demiştim.* Denklem homojen değilse özel çözümler vektör uzayı olmaz.*

Örneğin sabit katsayılı, 2. mert., lineer fakat homojen olmayan diferansiyel
denklemi ele alalım.
Ay''+By'+Cy=Q(x) olsun. (A, B, C, x ve y'nin fonksiyonları.)

y1 ve y2 özel çözümler ise y1+y2 bir çözüm değildir. Çünkü (Ay1''+By1'+Cy1)+
(Ay2''+By2'+Cy2)=Q(x)+Q(x)

A(y1''+y2'')+B(y1'+y2')+C(y1+y2)=2Q(x)

A(y1+y2)''+B(y1+y2)'+C(y1+y2)=2Q(x)


A(y1+y2)''+B(y1+y2)'+C(y1+y2)=2Q(x)=Q(x) ise yani Q sıfır ise, yani denklem
homojense vektör uzayıdır.
Yani, homojen olmayan denklemlerin özel çözümleri vektör uzayı olmuyor...

Ege

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