Download Lesson 11: Method of parameter variation In each of problems 1

Lesson 11: Method of parameter variation
In each of problems 1 through 6 use the method of variation of parameters to
find a particular solution of the given differential equation.
1. y 00 − 5y 0 + 6y = 2ex
2. y 00 − y 0 − 2y = 2e−x
3. y 00 + 2y 0 + y = 3e−x
4. 4y 00 − 4y 0 + y = 16ex/2
5. y 00 + y = tan x, 0 < x < π/2
6. y 00 + 4y 0 + 4y = x−2 e−2x , x > 0
Solution of 1. Let us first solve the homogeneous equation
y 00 − 5y 0 + 6y = 0.
The associate algebraic equation is
m2 − 5m + 6 = 0,
with roots m1 = 3, m2 = 2. Therefore, the solution of homogeneous equation is
c1 e3x + c2 e2x .
Let us now find the solution of the non-homogeneous equation as
y = f (x) = u1 (x)e3x + u2 (x)e2x .
where
Z
u1 (x) = c1 −
Z
u2 (x) = c2 +
y2 (x)g(x)
dx,
W [y1 (x), y2 (x)]
y1 (x)g(x)
dx.
W [y1 (x), y2 (x)]
In our case y1 (x) = e3x , y2 (x) = e2x , g(x) = 2ex , and
W [y1 (x), y2 (x)] = e3x 2e2x − e2x 3e3x = −e5x .
Therefore
Z
u1 (x) = c1 −
e2x 2ex
dx = c1 + 2
−e5x
Z
e3x 2ex
dx = c2 − 2
−e5x
Substituting it for y(x), we finally obtain
u2 (x) = c2 +
Z
e−2x dx = c1 − e−2x
Z
e−x dx = c2 + 2e−x .
y(x) = c1 e3x + c2 e2x + ex .
Solution of 4. We start from the homogeneous equation
1
y 00 − y 0 + y = 0
4
1
2
The associate algebraic equation is
1
= 0,
4
= 1/2. Therefore, the solution of homogeneous equation is
m2 − m +
with equal roots m1,2
c1 ex/2 + c2 xex/2 .
Let us now find the solution of the non-homogeneous equation as
y = f (x) = u1 (x)ex/2 + u2 (x)xex/2 .
where
Z
y2 (x)g(x)
dx,
W [y1 (x), y2 (x)]
Z
y1 (x)g(x)
dx.
u2 (x) = c2 +
W [y1 (x), y2 (x)]
In our case y1 (x) = ex/2 , y2 (x) = xex/2 , g(x) = 16ex/2 , and
´
³
1
1
W [y1 (x), y2 (x)] = ex/2 ex/2 + xex/2 − xex/2 ex/2 = ex .
2
2
Therefore
Z
xex/2 16ex/2
u1 (x) = c1 −
dx = c1 − 8x2 ,
ex
Z x/2 x/2
e 16e
u2 (x) = c2 +
dx = c2 + 16x.
ex
The final solution is
u1 (x) = c1 −
y(x) = c1 ex/2 + c2 xex/2 + 8x2 ex/2 .
E-mail address: vyachesl@inter.net.il