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Math 567 HW2 Due 9/
- Find the general solution, or the solution to the initial value problem, as appropriate
y′′^ 4 y x
Homogeneous: y′′^ 4 y 0 , ^2 4 0 , 2 i , y 1 cos 2x , y 2 sin 2x
Particular (by undetermined coefficients): yp Ax B
Lyp 4 Ax B x , 4A 1 , B 0 , yp 1 4
x
y c 1 cos 2x c 2 sin 2x 1 4
x
y′′^ 4 y′^ x , y 0 0 , y′ 0 0
Homogeneous: ^2 4 0 , 4 0 , 0, −4. yh c 1 c 2 e−^4 x
Particular (by undetermined coefficients): yp xAx B Ax^2 Bx
Lyp 2 A 4 2 Ax B x , 8A 1 , 2A 4 B 0 , A 1 8
, B − 1
, yp 1 8
x^2 − 1 16
x
y c 1 c 2 e−^4 x^ 1 8
x^2 − 1 16
x
y 0 0 c 1 c 2
y′ 0 0 − 4 c 2 − 1 16
c 2 − 1 64
, c 1 1 64
y 1 64
e−^4 x^ 1 8
x^2 − 1 16
x
y′′^ 3 y′^ − 4 y ex^ − e−x, y (^0) −1 , y′ (^0) 2
^2 3 − 4 0 , (^) (^4) − (^1) 0 , 1, − 4
yh c 1 ex^ c 2 e−^4 x
Particular (by undetermined coefficients):
Break up into
Prob 1 Prob 2
y′′^ 3 y′^ − 4 y ex^ y′′^ 3 y′^ − 4 y e−x
yp xAex^ yp Ae−x
Lyp 2 Aex^ 3 Aex^ ex^ , A 1 5
Lyp Ae−x^ − 3 e−x^ − 4 e−x^ − 6 Ae−x^ e−x^ ,
yp 1 1 5
xex^ yp 2 − 1 6
e−x
yp yp 1 − yp 2 1 5
xex^ − − 1 6
e−x^ 1 6
e−x^ 1 5
xex
y c 1 ex^ c 2 e−^4 x^ 1 6
e−x^ 1 5
xex^ General solution
y 0 − 1 c 1 c 2 1 6
y′ 0 2 c 1 − 4 c 2 − 1 6
c 1 − 27 50
, c 2 − 47 75
y − 27 50
ex^ − 47 75
e−^4 x^ 1 6
e−x^ 1 5
xex
y′′^ − 4 y′^ 5 y 0 , y 0 1 , y′ 0 2
^2 − 4 5 0 ,
2 i
y c 1 e^2 x^ cos x c 2 e^2 x^ sin x
y 0 c 1 1 , y′ 0 2 c 1 c 2 2 , c 2 0
y e^2 x^ cos x
- For the equation y′′^ − 4 y′^ 5 y 0
y c 1 e^2 x^ cos x c 2 e^2 x^ sin x
a) Find the solution y 1 satisfying y 0 1 , y′ 0 0
y 0 c 1 1 , y′ 0 2 c 1 c 2 0 , c 2 − 2 so y 1 e^2 x^ cos x − 2 e^2 x^ sin x
b) Find the solution y 2 satisfying y 0 0 , y′ 0 1
y 0 c 1 0 , y′ 0 2 c 1 c 2 1 , c 2 1 , so y 2 e^2 x^ sin x
c) How do y 1 and y 2 from parts a) and b) allow you to then solve the general initial value problem for this equation, i.e.
y′′^ − 4 y′^ 5 y 0 , y 0 a , y′ 0 b
The solution of this problem is then
y ay 1 by 2 ae^2 x^ cos x − 2 e^2 x^ sin x be^2 x^ sin x ae^2 x^ cos x − 2 a be^2 x^ sin x
(Terminology: The y 1 and y 2 calculated above are called the fundamental set of solutions
normalized at x 0. )
- Suppose the DE y′′^ pxy′^ qxy 0 has px, qx both continuous in an open interval
I containing x 0. Show that there is no choice of p, q
for which both y x and y sin 2x are solutions. (Hint: What initial conditions do these
functions satisfy at x 0? )
Via the hint: If we have y 1 x and y 2 sin 2x then y 1 0 0 , y 1 ′^ 0 1 and
y 2 0 0 , y 2 ′^ 0 2. Under the hypotheses of the problem if y 1 and y 2 are solutions that
satisfy these conditions then y 2 ≡ 2 y 1 since both sides would satisfy the same initial
conditions at x 0. Since clearly sin 2x ≠ 2 x these two functions cannot both be solutions of
the given DE under the hypotheses.
Alternately, we can argue that y 1 x and y 2 sin 2x cannot be a fundamental set of
solutions of the DE because W det
x sin 2x
1 2 cos 2x
2 x cos 2x − sin 2x and so W 0 0 but
Wx ≠ 0 for x in an interval about 0 and this can’t happen under the hypotheses of the
problem.
- We are given the constant coefficient homogeneous 2nd order linear DE,
Ly ay′′^ by′^ cy 0
a) If y x is a solution, show that ′x is also a solution.
We have a′′^ b′^ c 0. Taking the derivative, we obtain, a′′′^ b′′^ c′^ 0 so that,
upon rewriting, a′^
′′ b′^
′ c′^ 0. This shows that ′^ is a solution. b) Show that if y x is a solution that satisfies x 0 0 , ′x 0 1 then y ′x is
