解工程數學

Consider a(dy/dx)+by=k

if y1 and y2 is two solutions of it

Then
a(dy1/dx)+by1=k...(1)
a(dy2/dx)+by2=k...(2)

(1)-(2)
[a(dy1/dx)+by1]-[a(dy2/dx)+by2]=k-k=0
a(d/dx)(y1-y2)+b(y1-y2)=0

Shows that the difference is the solution of the homogeneous ODE

41
a(dy1/dx)+by1=k...(1)

a(dcy1/dx)+bcy1=kc

So, cy1 is a solution of the non-homogeneous ODE equation where the constant term is replaced by c times the original constant

42

(dy1/dx)+py1=r1; (dy2/dx)+py2=r2

(dy1/dx)+py1+ (dy2/dx)+py2=r1+r2

(d/dx)(y1+y2)+p(y1+y2)=r1+r2

We conclude that y1+y2 is a solution of the ODE

dy/dx+py=r1+r2