Transcript Bのcos

Nonhomogeneous
Linear Differential
Equations
AP CALCULUS BC
Nonhomogeneous Differential Equations
Recall that second order linear differential equations with constant
coefficients have the form:
ay  by  cy  G ( x)
Now we will solve equations where G(x) ≠ 0, which are
non-homogeneous differential equations.
Complementary Equation
The related homogeneous equation ay  by  cy  0 is called the
complementary equation, and it is part of the general solution to the
nonhomogeneous equation.
General Solution
The general solution to a nonhomogenous differential equation is
y(x) = yp(x) + yc(x)
where yp is a particular solution to the nonhomogenous equation, and yc
is the general solution to the complementary equation.
Proof for Grant
yc(x) = y(x) – yp(x)
 a ( y  y p )  b( y  y p )  c( y  y p )
 ay  ay p  by  by p  cy  cy p
 (ay  by  cy )  (ay p  by p  cy p )
 g ( x)  g ( x)  0
Method of Undetermined Coefficients
There are two methods for solving nonhomogeneous equations:
Method of Undetermined Coefficients
Variation of Parameters
First, we will learn about the Method of Undetermined Coefficients to
solve the equation ayʹʹ + byʹ + cy = G(x) when G(x) is a polynomial.
Since G(x) is a polynomial, yp is also a polynomial of the same degree as G.
Therefore, we substitute yp = a polynomial (of the same degree as G) into
the equation and determine the coefficients.
Example 1
Solve the equation yʹʹ + yʹ – 2y = x2.
The auxiliary equation is r2 + r – 2 = 0
Factor  (r – 1)(r + 2) = 0
r = 1, r = –2
Solution of complementary equation is yc = c1ex + c2e–2x
Since G(x) = x2, we want a particular solution where
yp(x) = Ax2 + Bx + C
So ypʹ(x) = 2Ax + B and ypʹʹ(x) = 2A
Example 1 (continued)
Substitute these into the differential equation 
(2A) + (2Ax + B) – 2(Ax2 + Bx + C) = x2
= – 2Ax2 + (2A – 2B)x + (2A + B – 2C) = 1x2 + 0x + 0
1
–2A = 1  A  
2
 1
2A – 2B = 0  2     2 B  0
 2
1
B
2
2 A  B  2C  0
 1  1
2         2C  0
 2  2
3
C
4
Example 1 (FINAL)
Therefore, our particular solution is
1 2 1
3
yp   x  x 
2
2
4
So our general solution is:
ex
y = yc + yp = c1 +
1
–2x
c2e 
1
3
x  x
2
2
4
2
Example 2
Solve yʹʹ + 4y = e3x
When G(x) is of the form ekx, we use yp = Aekx because the
derivatives of ekx are constant multiples of ekx and work out nicely.
Complementary equation is r2 + 4 = 0  r  2i
Therefore, yc = c1 cos 2x + c2 sin 2x
Solve for yp:
y = Ae3x
yʹ = 3Ae3x
yʹʹ = 9Ae3x
Substitute  9 Ae3 x  4 Ae3 x  e3 x
Example 2 (continued)
9 Ae3 x  4 Ae3 x  e3 x
13 Ae3 x  e3 x
13A = 1
A = 1/13
1 3x
yp  e
13
1 3x
General solution is y  c1 cos 2 x  c2 sin 2 x  e
13
Example 3
Solve the equation yʹʹ + yʹ – 2y = sin x.
When G(x) is of the form C sin kx or C cos kx, we use
yp = A cos kx + B sin kx
Complementary equation is r2 + r – 2 = 0  (r  2)(r  1)  0
x
2 x
y

c
e

c
e
r = –2, 1  c 1
2
yp = A cos x + B sin x
ypʹ = –A sin x + B cos x
ypʹʹ = –A cos x – B sin x
Example 3 (continued)
Substitute back into original equation:
(–A cos x – B sin x) + (–A sin x + B cos x) – 2(A cos x + B sin x) = sin x
(–3A + B) cos x + (–A – 3B) sin x = sin x
Therefore, –3A + B = 0 and –A – 3B = 1
1
3
Solve as a system  A  
B
10
10
1
3
y p   cos x  sin x
10
10
1
3
x
2 x
General solution is y  c1e  c2e  cos x  sin x
10
10
Review and More Rules for
Method of Undetermined Coefficients
Form is ayʹʹ + byʹ + cy = G(x)
1.
If G(x) is a polynomial, use yp = Axn + Bxn–1 + … + C.
2.
If G(x) = Cekx, use yp = Aekx .
3.
If G(x) = C sin kx or C cos kx, use yp = A cos kx + B sin kx
4.
If G(x) is a product of functions, multiply them for yp.
Example: G(x) = x cos 3x  yp = (Ax + B) cos 3x + (Cx + D) sin 3x
5.
If G(x) is a sum of functions, find separate particular solutions
and add them together at the end.
Example: G(x) = xex + cos 2x
Use yp1 = (Ax + B)ex and yp2 = C cos 2x + D sin 2x
Then add  y(x) = yc + yp1 + yp2
6.
If yp is a solution to the complementary equation (yc), multiply yp
by x or x2, so yc and yp are linearly independent.
7.
The particular solutions we try to find using yp (the ones with the A,
B, C, etc. in them) are called “trial solutions”.