MATH3-Lect9.pdffsasasasasasasasasdcldldlc

maths diffrential

1. MATHEMATICS-III
Lecture-9
P. DANUMJAYA
Department of Mathematics
BITS-Pilani K K Birla Goa Campus
Finding Particular Solution P. Danumjaya

2. Finding Particular Solution
by using
The Method of Undetermined Coefficients
Finding Particular Solution P. Danumjaya

3. Theorem 1
Ifygis the general solution of the differential equation
y
′′
+P(x)y
′
+Q(x)y=0, (1)
andypis any particular solution of
y
′′
+P(x)y
′
+Q(x)y=R(x), (2)
thenyg+ypis the general solution of (2).
Finding Particular Solution P. Danumjaya

4. We consider the linear second order nonhomogeneous
differential equation
y
′′
+a y
′
+b y=f(x), (3)
wherea,bare constants andf(x)has the form:
f(x) =e
αx
f(x) = αxorcosαxorAsinαx+Bcosβx
f(x) =a0+a1x+· · ·+anx
n
.
Finding Particular Solution P. Danumjaya

5. Case(i)f(x) =e
αx
Assume thatf(x)has the formf(x) =e
αx
.
Then we have
y
′′
+a y
′
+b y=e
αx
.
Letyp=Ae
αx
then substituting in the given equation,
we obtain
A=
1
α
2
+aα+b
.
Finding Particular Solution P. Danumjaya

6. Note
Ifαis a
m
2
+am+b=0,
then we choose
yp=A x e
αx
.
Ifαis a
m
2
+am+b=0,
then we choose
yp=A x
2
e
αx
.
Finding Particular Solution P. Danumjaya

7. Example 1
Find the general solution of
y
′′
−y=e
3x
.
The general solution is
y(x) =yg(x) +yp(x) =C1e
−x
+C2e
x
+
1
8
e
3x
,
whereC1andC2are any arbitrary constants.
Finding Particular Solution P. Danumjaya

8. Example 2
Find the general solution of
y
′′
+3y
′
−4y=e
−4x
.
The general solution is
y(x) =yg(x) +yp(x) =C1e
x
+C2e
−4x
−
x
5
e
−4x
,
whereC1andC2are any arbitrary constants.
Finding Particular Solution P. Danumjaya

9. Case(ii)f(x) = sinαx(or)cosαx
Suppose
y
′′
+a y
′
+b y= sinαx(or)cosαx(or)Msinαx+Ncosαx,
then we choose the particular solution as
yp=Asinαx+Bcosαx.
Ifypsatisfies the homogeneous equation
y
′′
+a y
′
+b y=0,
then we need to choose
yp=x(Asinαx+Bcosαx).
Finding Particular Solution P. Danumjaya

10. Case(ii)f(x) = sinαx(or)cosαx
Suppose
y
′′
+a y
′
+b y= sinαx
(or)cosαx(or)Msinαx+Ncosαx,then we choose the particular solution as
yp=Asinαx+Bcosαx.
Ifypsatisfies the homogeneous equation
y
′′
+a y
′
+b y=0,
then we need to choose
yp=x(Asinαx+Bcosαx).
Finding Particular Solution P. Danumjaya

11. Case(ii)f(x) = sinαx(or)cosαx
Suppose
y
′′
+a y
′
+b y= sinαx
(or)cosαx(or)Msinαx+Ncosαx,then we choose the particular solution as
yp=Asinαx+Bcosαx.
Ifypsatisfies the homogeneous equation
y
′′
+a y
′
+b y=0,
then we need to choose
yp=x(Asinαx+Bcosαx).
Finding Particular Solution P. Danumjaya

12. Example 3
Find a particular solution of
y
′′
+4y= cosx,
and hence find a general solution.
Finding Particular Solution P. Danumjaya

13. Solution
Let
yp(x) =Acosx+Bsinx.
Substituting in the given equation, we find
yp(x) =
1
3
cosx.
The general solution is
y(x) =yg(x) +yp(x) =C1cos2x+C2sin2x+
1
3
cosx,
whereC1andC2are any arbitrary constants.
Finding Particular Solution P. Danumjaya

14. Example 4
Find the particular solution of
y
′′
+4y= cos2x.
Finding Particular Solution P. Danumjaya

15. Solution
Let
yp(x) =x(Acos2x+Bsin2x).
Substituting in the given equation, we find
yp(x) =
x
4
sin2x.
Finding Particular Solution P. Danumjaya

16. Case(iii)f(x) =a0+a1x+a2x
2
+···+anx
n
If
y
′′
+a y
′
+b y=a0+a1x+a2x
2
+· · ·+anx
n
,
then we choose the particular solution as
yp=A0+A1x+A2x
2
+· · ·+Anx
n
.
Note that ifb=0 x
n−1
as the
highest power ofx.
In this case, we chooseypas
yp=x
ˇ
A0+A1x+A2x
2
+· · ·+Anx
n
ı
.
Finding Particular Solution P. Danumjaya

17. Example 5
Find the general solution of
y
′′
−y=x
2
+1.
Finding Particular Solution P. Danumjaya

18. Solution
Let
yp(x) =A0+A1x+A2x
2
.
Substituting in the given equation, we find
yp(x) =−x
2
−3.
The general solution is
y(x) =yg(x) +yp(x) =C1e
x
+C2e
−x
−x
2
−3,
whereC1andC2are any arbitrary constants.
Finding Particular Solution P. Danumjaya

19. Theorem 2
Ify1(x)andy2(x)are solutions of
y
′′
+P(x)y
′
+Q(x)y=f1(x),
and
y
′′
+P(x)y
′
+Q(x)y=f2(x),
respectively, theny1(x) +y2(x)is a solution of
y
′′
+P(x)y
′
+Q(x)y=f1(x) +f2(x).
Finding Particular Solution P. Danumjaya

20. Example 6
Find the general solution of
y
′′
+4y=4cos2x+6cosx+8x
2
−4x.
Finding Particular Solution P. Danumjaya

21. Solution
The general solution is
y(x) =yg(x) +yp(x) =C1cos2x+C2sin2x+xsin2x+2cosx
−1−x+2x
2
,
whereC1andC2are any arbitrary constants.
Finding Particular Solution P. Danumjaya

22. THANK YOU !
Finding Particular Solution P. Danumjaya