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WorksheetsSecond Order Linear DE
Total questions: 10
Worksheet time: 5mins
Standard form for 2nd order linear DE is
ay′′+by′+cy=0
ay′′+by′+cy=Q
y′′+by′+cy=0
y′′+by′+cy=Q
A 2nd order linear DE ay′′+by′+cy=Q is homogeneous if
a=0
b=0
c=0
Q=0
A 2nd order linear DE ay′′+by′+cy=Q has a general solution of the form
y=yc+yp
y=C1eαx+C2eβx
Iy=∫IQ dx
y=(C1+C2x)eαx
To find the complementary function yc of the general solution of ay′′+by′+cy=Q we use characteristic equation
am2+bm+c=Q
am2+bm+c=0
m2+bm+c=0
am2+bm=c
A 2nd order linear DE ay′′+by′+cy=Q can be written in the D-operator form as
aD2+bD+c=Q
aD2y+bDy+cy=0
aD2y+bDy+cy=Q
aD2y+bDy+c=Q
The D-operator Dy is equivalent to
dx2d2y
dxdy
y′′
y′
The D-operator D2y is equivalent to
dx2d2y
dxdy
y′′
y′
To find the particular function yp of the general solution of ay′′+by′+cy=Q we use inverse D-operator
yp=(aD2+bD+c)(Q)
yp=D1(Q)
yp=aD2+bD+c1(Q)
yp=aD2+bD+c1(0)
The particular function yp of y′′+2y′+y=5 is
yp=(D2+2D+1)(5)
yp=D1(5)
yp=D2+2D+11(5)
yp=D2+2D1(5)
The inverse D-operator D1[f(x)] is equivalent to
∫f(x) dx
dxd(f(x))
f′(x)
∫f′(x) dx
